Theres a mistake here at around 10:38 that would be catastrophic if you were summing to n instead of infinity. You should be taking the limit of (1+ 1/2 - 1/n - 1/{n+1}) - the 1/n you missed comes from the (n-1)th bracket...
The key difference to take home between the integral and the summation is that the integral is summing up infinitely many real values of x, of which there are infinitely many more infinities thrown in to the mix, but the summation is summing up only the integer values of x from 2 to infinity. The summation converges BECAUSE the integral converges, because the summation is adding up fewer values of x, even though in both cases you're still going off to infinity regardless. The cardinality of the set of real numbers is infinitely infinitely greater than the cardinality of the set of the integers, because between any two integer values of x, there are infinitely many real values of x. This is why the reals are said to be dense over the integers. The weird paradox in all of this though, is that ln(3) is approximately 1.1 whereas 3/2 is precisely 1.5. So the integral converges to a smaller value than the summation. An absolutely bizarre result, and not one that I myself understand. Anybody else willing to chime in and help me (and others) learn why? :)
You can imagine summation as adding the areas of rectangles, each of which has f(x) * 1 dimensions. This will be larger than the area under the curve (i.e. the integral), because of the parts of the rectangles "going beyond" the curve.
Just learn definition of derivative and proof of the fundemental theorem. You should be caught up and you should able to prove everything else by your self.
This is one of the most interesting videos of yours. Why does it converge to different values? Value for integral is ln 3 = 1.09 and for Summation is 1.5 .They are very near to each other.
Quahntasy - Animating Universe I think because when you use the integral method you’re inputting non-integers as well, so the answer you get isn’t what the series converges to, but simply an estimate to tell you whether the series converges or not,,, I might be wrong tho
There's perhaps a more direct way to show the two parts of the summation are telescoping, and that is by re-indexing. The first term is a harmonic series starting with index n = 1, and the second term is a harmonic series starting at index n = 3. Then you can match the two term by term and show which parts of the two series (subtracting the second from the first) cancel directly by inspection. All that are left art the terms from the first series with n=1 and n = 2. Easy-peasy and not even cheesy. Apologies to anyone who already showed this; too many comments to read through.
Both the integral and the series are technically just a series. The series is a series with a multiplier of Δx = 1, whereas the integral is the same expression with Δx -> 0. It would be interesting to calculate the expression for arbitrary Δx, and then the integral and the series are just especial values of the expression, which would be a function of Δx. Am I onto something? Yes! This is, I’m introducing time-scale calculus to this baby!
David vd Lugt My understanding is that we need to use generalized shifting operators to be able to solve it, but otherwise, it is not much different than generally solving a difference equation. However, that in itself is quite challenging, as it is typically never a simple difference equation.
In principle one can apply the telescoping sum trick to the integral: 2:03 At this point one can do a simple variable substitution x-> x+-1 to get Integral over 1/x dx from 1 to infinity minus Integral over 1/x from 3 to infinity. I.e. same integral subtracted from each other over different x-Range. Result is Integral over 1/x dx from 1 to 3. Again the infinite part vanishes.
I might be wrong, but the lim when x->inf of ln( (x-1)/(x+1) ) is undetermined, you can't just divide it. However, using the L'Hopital method it is ln( 1/1) which is equal to 0.
For telescoping series, it seems it would be easier to rewrite the two series after re-defining "n" as the difference between the sum from 1 to inf and the sum from 3 to inf. All except for 1st two terms in 1 to inf series cancel. They are 1 & 1/2. Done.
I memorized how to calculate the inverses for hyperbolic tangent and hyperbolic cotangent and now i feel ultimately powerful 😂 It's actually easier to memorize them together since the way the x's are set up in the numerator and denom are just flipped
can you do a video of how to convert integrals to sums and vice versa? (i've done it in physics but not rigorously, when you go from a discrete situation to the analogous continuous situation and you have a delta x tend to zero and turn into dx). This LOOKS like i a case of this but the dx actually matters and if the sum was n/(n^2-1) it would be closer to analogous although i don't think that's the actual conversion.
"... but the dx actually matters and if the sum was n/(n^2-1) it would be closer to analogous ..." No, actually, the dx is a sort of stand-in for the summation itself. Consider, e.g., ∫ 1 dx vs ∑ 1 Fred
@@ffggddss but consider the units. if you have sum x vs. integral of x dx , the first has units of x whereas the second has units of x^2. , or e.g. displacement = velocity * time when doing averages or discretely etc., but when you integrate the equivalent is integral v dt
This video was awesome! Buddy, i would like you to show how we can solve polinomial equations including trigonometrical functions, like ax^2+sin(x). Thanks.
Why did you do partial fractions for the integral? You could've simply realized it is -arctanh and did a simple factoring to get that it's -2arctanh(x)+C. Then from there you could've referenced the definition of arctanh(x) from either memory or an online resource to figure out for that domain it is ln(1-x)-ln(1+x)+C. Which would be far less Partial Fractions..
Very nice video! I'd like to show another way to do the infinite sum of 1/(n-1) and -1/(n+1). You can split them up into sum[1/(n-1)] - sum[1/(n+1)]. Then do a "change of variables" in one of the sums to convert them into the other. For example, change the 1/(n-1) to 1/(n+1), but let n start from 0 instead of 2 so that it'll be in the same form as the other sum. Then write down the n=0 and n=1 term explicitly, and you end up with 1/1 + 1/2 + sum[1/(n+1)] - sum[1/(n+1)], with n starting from 2 for both sums. The sums cancel out, and you get 3/2 as the final answer. This also works if you change the 1/(n+1) sum to 1/(n-1) with n starting from a different integer. This lets you avoid having to write down several terms hoping to find a pattern where they cancel :)
That technique doesn't work in general, although it's fine in this case. Each of the sums you broke it into diverges, which messes everything up. The problem is that, if you consider 1/1-1/3+1/2-1/4+1/3-1/5... as a single series with alternating hard-to-generalize terms, it doesn't converge absolutely, so you can't rearrange the terms arbitrarily. What you're ultimately trying to do ends up working, though, and I think it's because the series converges and you're only moving each term a bounded distance between the original and where you end up.
@@iabervon Thank you, I did not consider the individual sums to be divergent. But in this case, the original series does converge absolutely. The absolute value of 1/(n-1) - 1/(n+1) is itself for n>=2, since 1/(n-1) > 1/(n+1) for n>=2, so their difference is always positive. And by the integral test as shown in BPRP's video, the original series converges. Because the summand in this case is equal to its absolute value in the given interval, then the series also converges absolutely, allowing us to rearrange the individual terms arbitrarily. Maybe if a series is absolutely convergent, and if it can also be written as a sum of divergent series, then we are allowed to manipulate those divergent series arbitrarily without changing the sum's value?
Nice video! But I wonder why it is possible to remove the brackets from this infinite sum? Should it not be proven before that the sum is absolutely convergent?
A couple of remarks: 1) Is there any relationship between the integral and infinite series that we can draw? For example, will the infinite series always be greater in value than the improper integral? 2) Is it correct that the difference between the two is that the integral is because it is in the continuous world, whereas the infinite series is in the discrete world? For example, discrete probability has to do with various values, such as numbers on dice, whereas continuous probability (continuous = use of calculus and use of integrals). Can we draw these conclusions? 3) In addition, I was wondering if there is any way which we can determine what the value of infinite series which are not conventional is - for example, something similar to a Simpson’s Rule or other approximation technique which works on integrals - but for infinite series? Thanks, ~J.J.
For 1) and 2), you can compare the sum & the integral more faithfully by depicting the sum as the integral of a stair-step function whose "steps" are the values of the summand, carried as constants, from n to n+1. So when, like here, the function being summed/integrated is monotonic decreasing, the sum will have to be > the integral. If the function were monotonic increasing, the sum would be < the integral. This is of course, for the case where the limits of summation and integration are the same. For 3), OK, Simpson's Rule is a way of numerically approximating an integral. Generally, you can improve the accuracy by densifying the intervals. If you're looking for an analogy for infinite series, I suppose you could look at the various techniques for accelerating convergence of a series. It's not a short topic; there's no single answer. There are a few that work in large classes of cases - term-averaging, e.g. - but it's a lot like differential equations - you rummage around the toolbox looking for something that works. Fred
Curious why you choose to hold a clunky ball microphone rather than clipping one on? ... especially since it looks like you have recently replaced it with a smaller version. You could use your left hand for the red pen. ;^))
8:55 by cancelling terms you must switch the order of the these terms, infinity times. You split every term into 2 terms, with one > 0 and another one < 0, then the series only converges conditionally, you can't change the order of the terms to make them cancel to each other. en.wikipedia.org/wiki/Conditional_convergence
Can you explain why theta substitution gives a different answer? I am putting x=sec(theta) and we get integral of cosec(theta), which isn't the same as the partial fraction answer. ???
Hey if the integral limit would be 0 to inf. Then what were the integral result. I mean to say that 1/0->inf. But when we split it 0 to 2 and 2 to inf. Then can be solved. But....... Try it like u did for 2 to inf.
How does the sum of the series come out to less than that of the integral? I always thought that the integral had a higher value, which was why the integral convergence test was valid.
Both are decreasing with increasing x (or n). The sum "holds" the value until the next term, while the integral is decreasing. Just sketch the graphs and you'll see that the step function representing the sum is "on top of" the continuous function.
@@pjmmccann you can draw the series as a rectangles with width equal to 1 and hight equal to the function but if you start with a rectangle from the left you will see rectangles is less than the continous area so if you substract from the series [(3/2) - 1/(4-1)] you get (7/6) which is still bigger than ln(3)
You do not even need to limit the 1/(n+1) you dont even need to mention it, because it will be cancelled out by the next-next term. Just as all the other terms except for 1+1/2
That's not quite true. Consider the sum of (n - (n - 1)) from n = 2 to infinity. The terms all cancel except for -1, but the series is 1+1+1+... which obviously shouldn't be -1. The reason it doesn't work in this case is that the leftover term that you're going to cancel later doesn't go to 0 as n goes to infinity.
@@iabervon you are describing a different PROPER sum. This is an IMPROPER telescopic sum, meaning all the terms except for a finite number of them (in the front) eventually cancel out (after the achemedic principle "for every natural number n there is a higher number n+1") . Due to the archimedic principle it only matters what happens with terms of infinite index, not those of finite index.
Is there a function where the infinite sum and the definite integral from the same number to infinity are equal? No floor function tho, that’s cheating lol
You can graph the sum as 1-wide rectangles whose height matches the curve at the left. If the function is decreasing everywhere, that's obviously bigger.You could probably find a sin(kn+m)/n that worked, though.
I forgot one term in the telescoping series, please see correction here: ua-cam.com/video/CLthG-tZkpU/v-deo.html
4:13 Never in my life have I seen such a beautifully and perfectly written 1
Л.С. Мото lol glad to hear!!
(Well well) * ( isn’t it isn’t it )
Jake Andrews lolll love it!!!
Can someone explain this pls
@@dimes7742 nope
@@dimes7742 (1+1)(-1-1)
Im the 100th like
Theres a mistake here at around 10:38 that would be catastrophic if you were summing to n instead of infinity. You should be taking the limit of (1+ 1/2 - 1/n - 1/{n+1}) - the 1/n you missed comes from the (n-1)th bracket...
(1 + 1/2 - 1/n - 1/(n+1))*
@@stephenphelps920 Thats what I meant, my bad
Was literally about to comment the same thing , though it would be zero but an error is an error
He demonstrated earlier that the (n-1) would cancel out with a factor that came before. You don’t include it in the limit.
Not quite. The 1/n term is cancelled out from a previous term.
The key difference to take home between the integral and the summation is that the integral is summing up infinitely many real values of x, of which there are infinitely many more infinities thrown in to the mix, but the summation is summing up only the integer values of x from 2 to infinity. The summation converges BECAUSE the integral converges, because the summation is adding up fewer values of x, even though in both cases you're still going off to infinity regardless. The cardinality of the set of real numbers is infinitely infinitely greater than the cardinality of the set of the integers, because between any two integer values of x, there are infinitely many real values of x. This is why the reals are said to be dense over the integers.
The weird paradox in all of this though, is that ln(3) is approximately 1.1 whereas 3/2 is precisely 1.5. So the integral converges to a smaller value than the summation. An absolutely bizarre result, and not one that I myself understand. Anybody else willing to chime in and help me (and others) learn why? :)
Probably some negative areas did that
Same question pleas anybody reply
@@tipoima no from 2 to infinity
1/(x^2 -1) is positive
You can imagine summation as adding the areas of rectangles, each of which has f(x) * 1 dimensions. This will be larger than the area under the curve (i.e. the integral), because of the parts of the rectangles "going beyond" the curve.
Don’t know why I watch these videos because I haven’t took calculus yet but I do
Because numbers are pretty, i say
Just learn definition of derivative and proof of the fundemental theorem. You should be caught up and you should able to prove everything else by your self.
I haven't taken calculus either, but I feel like I won't have to if I keep watching these!
@@emperorpingusmathchannel5365 not a good idea, just take a damn calculus course
it seems you also haven't taken english yet xD
Love the "well well" at 3:31
karan jai singh sandhu thank you!!
Thanks man, this is the best math channel I've ever seen. this video helped me solve a problem that i had no idea where to start
Fascinating! Man that was a really good one. Really helped clear up my understanding.
This is one of the most interesting videos of yours. Why does it converge to different values?
Value for integral is ln 3 = 1.09 and for Summation is 1.5 .They are very near to each other.
Quahntasy - Animating Universe I think because when you use the integral method you’re inputting non-integers as well, so the answer you get isn’t what the series converges to, but simply an estimate to tell you whether the series converges or not,,, I might be wrong tho
That's a great question. Superb!!! This question exercises a lot of different mathematical techniques!!!
Great video! Why don't you make some videos on problems of math competions?
Pierangelo Errico maybe during my break.
There's perhaps a more direct way to show the two parts of the summation are telescoping, and that is by re-indexing. The first term is a harmonic series starting with index n = 1, and the second term is a harmonic series starting at index n = 3. Then you can match the two term by term and show which parts of the two series (subtracting the second from the first) cancel directly by inspection. All that are left art the terms from the first series with n=1 and n = 2. Easy-peasy and not even cheesy. Apologies to anyone who already showed this; too many comments to read through.
Both the integral and the series are technically just a series. The series is a series with a multiplier of Δx = 1, whereas the integral is the same expression with Δx -> 0. It would be interesting to calculate the expression for arbitrary Δx, and then the integral and the series are just especial values of the expression, which would be a function of Δx.
Am I onto something? Yes! This is, I’m introducing time-scale calculus to this baby!
Wow, never heard of this idea, sounds good
Can you explain how to start tackling that kind of problem
David vd Lugt My understanding is that we need to use generalized shifting operators to be able to solve it, but otherwise, it is not much different than generally solving a difference equation. However, that in itself is quite challenging, as it is typically never a simple difference equation.
In principle one can apply the telescoping sum trick to the integral:
2:03 At this point one can do a simple variable substitution x-> x+-1 to get
Integral over 1/x dx from 1 to infinity minus Integral over 1/x from 3 to infinity.
I.e. same integral subtracted from each other over different x-Range.
Result is Integral over 1/x dx from 1 to 3.
Again the infinite part vanishes.
Wow! Was an amazing demonstration. Every day I keep learning with this incredible mathematician.
What did we get when BPRP dug two holes looking for water? Well, well.
I might be wrong, but the lim when x->inf of ln( (x-1)/(x+1) ) is undetermined, you can't just divide it. However, using the L'Hopital method it is ln( 1/1) which is equal to 0.
For telescoping series, it seems it would be easier to rewrite the two series after re-defining "n" as the difference between the sum from 1 to inf and the sum from 3 to inf. All except for 1st two terms in 1 to inf series cancel. They are 1 & 1/2. Done.
I memorized how to calculate the inverses for hyperbolic tangent and hyperbolic cotangent and now i feel ultimately powerful 😂 It's actually easier to memorize them together since the way the x's are set up in the numerator and denom are just flipped
Please make a video about converting summation forms & series to integrals and vice-versa..
can you do a video of how to convert integrals to sums and vice versa? (i've done it in physics but not rigorously, when you go from a discrete situation to the analogous continuous situation and you have a delta x tend to zero and turn into dx).
This LOOKS like i a case of this but the dx actually matters and if the sum was n/(n^2-1) it would be closer to analogous although i don't think that's the actual conversion.
"... but the dx actually matters and if the sum was n/(n^2-1) it would be closer to analogous ..."
No, actually, the dx is a sort of stand-in for the summation itself. Consider, e.g.,
∫ 1 dx vs ∑ 1
Fred
@@ffggddss but consider the units. if you have sum x vs. integral of x dx , the first has units of x whereas the second has units of x^2. , or e.g. displacement = velocity * time when doing averages or discretely etc., but when you integrate the equivalent is integral v dt
@@khajiit92 x is unitless in this case.
You can see this by noting that the limits of integration are unitless.
Fred
Keep on going! Great videos!
This video was awesome! Buddy, i would like you to show how we can solve polinomial equations including trigonometrical functions, like ax^2+sin(x). Thanks.
OK, the integral is ln3; the sum is 3/2.
Both using the partial-fraction decomposition, 2/(x²-1) = 1/(x-1) + 1/(x+1)
Fred
Despite the harmless mistake people are talking about, the math still stands, so nice video!
Why did you do partial fractions for the integral? You could've simply realized it is -arctanh and did a simple factoring to get that it's -2arctanh(x)+C. Then from there you could've referenced the definition of arctanh(x) from either memory or an online resource to figure out for that domain it is ln(1-x)-ln(1+x)+C. Which would be far less Partial Fractions..
_Well, well_
Very nice video! I'd like to show another way to do the infinite sum of 1/(n-1) and -1/(n+1). You can split them up into sum[1/(n-1)] - sum[1/(n+1)]. Then do a "change of variables" in one of the sums to convert them into the other. For example, change the 1/(n-1) to 1/(n+1), but let n start from 0 instead of 2 so that it'll be in the same form as the other sum. Then write down the n=0 and n=1 term explicitly, and you end up with 1/1 + 1/2 + sum[1/(n+1)] - sum[1/(n+1)], with n starting from 2 for both sums. The sums cancel out, and you get 3/2 as the final answer. This also works if you change the 1/(n+1) sum to 1/(n-1) with n starting from a different integer. This lets you avoid having to write down several terms hoping to find a pattern where they cancel :)
That technique doesn't work in general, although it's fine in this case. Each of the sums you broke it into diverges, which messes everything up. The problem is that, if you consider 1/1-1/3+1/2-1/4+1/3-1/5... as a single series with alternating hard-to-generalize terms, it doesn't converge absolutely, so you can't rearrange the terms arbitrarily. What you're ultimately trying to do ends up working, though, and I think it's because the series converges and you're only moving each term a bounded distance between the original and where you end up.
@@iabervon Thank you, I did not consider the individual sums to be divergent. But in this case, the original series does converge absolutely. The absolute value of 1/(n-1) - 1/(n+1) is itself for n>=2, since 1/(n-1) > 1/(n+1) for n>=2, so their difference is always positive. And by the integral test as shown in BPRP's video, the original series converges. Because the summand in this case is equal to its absolute value in the given interval, then the series also converges absolutely, allowing us to rearrange the individual terms arbitrarily. Maybe if a series is absolutely convergent, and if it can also be written as a sum of divergent series, then we are allowed to manipulate those divergent series arbitrarily without changing the sum's value?
great video. i thought that the integral would be bigger than the power sum, i'm new to this xD
Nice video! But I wonder why it is possible to remove the brackets from this infinite sum? Should it not be proven before that the sum is absolutely convergent?
@VeryEvilPettingZoo That was a really nice answer. Thank you very much!
お見事です
うーん 因数分解なんて何の役に立つんだーってひとはこれ見て欲しい！
そこでサボるとこれが解けないもんね！
It was never this much fun when I knew there was going to be a test over it.
Minor item. The summation should be done as i = 2 to n (not n = 2 to n) and then watch as n goes to infinity.
for x^(-x), the integral from 0 to 1 and the series from 0 to inf are equal
A couple of remarks:
1) Is there any relationship between the integral and infinite series that we can draw? For example, will the infinite series always be greater in value than the improper integral?
2) Is it correct that the difference between the two is that the integral is because it is in the continuous world, whereas the infinite series is in the discrete world? For example, discrete probability has to do with various values, such as numbers on dice, whereas continuous probability (continuous = use of calculus and use of integrals). Can we draw these conclusions?
3) In addition, I was wondering if there is any way which we can determine what the value of infinite series which are not conventional is - for example, something similar to a Simpson’s Rule or other approximation technique which works on integrals - but for infinite series?
Thanks,
~J.J.
For 1) and 2), you can compare the sum & the integral more faithfully by depicting the sum as the integral of a stair-step function whose "steps" are the values of the summand, carried as constants, from n to n+1.
So when, like here, the function being summed/integrated is monotonic decreasing, the sum will have to be > the integral.
If the function were monotonic increasing, the sum would be < the integral.
This is of course, for the case where the limits of summation and integration are the same.
For 3), OK, Simpson's Rule is a way of numerically approximating an integral. Generally, you can improve the accuracy by densifying the intervals.
If you're looking for an analogy for infinite series, I suppose you could look at the various techniques for accelerating convergence of a series. It's not a short topic; there's no single answer. There are a few that work in large classes of cases - term-averaging, e.g. - but it's a lot like differential equations - you rummage around the toolbox looking for something that works.
Fred
Curious why you choose to hold a clunky ball microphone rather than clipping one on? ... especially since it looks like you have recently replaced it with a smaller version. You could use your left hand for the red pen. ;^))
(Havent done anything close to this math yet)
Why is - 1/(n-1) a surviving term? Will that not just be canceled by a term if you continue the series?
8:55 by cancelling terms you must switch the order of the these terms, infinity times. You split every term into 2 terms, with one > 0 and another one < 0, then the series only converges conditionally, you can't change the order of the terms to make them cancel to each other. en.wikipedia.org/wiki/Conditional_convergence
Plzz integrate
ln(ln(ln(lnx)))
Gourav Madhwal instant like!
math is fun. good video. Thank you.
The reason the integral is less than the sum is because the sum is an integral of the floor of x which is less than x itself :)
really liked the telescopic series,
Can u show us examples where the integral equals the sommation
Muy interesante esa comparación.
Please make a video on how to convert summation series into integrals.
Can you explain why theta substitution gives a different answer?
I am putting x=sec(theta) and we get integral of cosec(theta), which isn't the same as the partial fraction answer.
???
Amazing!
Hey BpRp, which calculus texts do you most recommend? Thanks!
This is so cool
If you sobstitute floor(x) to x in the integral they are the same!
О привет, интегральный признак сходимости Коши. Я бы даже сказал здравствуй
I prefer the integral !
Could you make a video converting Infinite sums to Integrals and vice versa?
Lim p->YAY (bprp)=(YAY)!
I don't know why but infinite series is more appealing than the integral for me
@@Gameboygenius 🐒🤔do astronomers love infinite series more
@@karthikrambhatla7465 I think that's just a plain pun around telescopes.
@@Gameboygenius lol!! Got you 😂
Hey if the integral limit would be 0 to inf. Then what were the integral
result.
I mean to say that 1/0->inf.
But when we split it 0 to 2 and 2 to
inf. Then can be solved. But.......
Try it like u did for 2 to inf.
something that i can do by myself :)))
Why do I watch this channel. I'm in algebra 1 . Lol
Noooo, I worked out the solution to the thumbnail but the question in the video was slightly different :(
OMG, I am sorry.
Please multiply your answers by 2 to solve the problems.
I think I did well on my calc 2 final
so for the series on the right if i were to stop at n=11 the value would be (19/12 1,5+1/12)? just to get it right
What about substituting the variable x with 'tan'
Yay
Did you use the Residue definition to evaluate the partial fraction decomposition?
Mr Breiart
No. It's just the "cover up method" for partial fractions. I have vids on that.
@@blackpenredpen I'm checking them out then, thanks :)
🎉good
Make a video on partial fraction....😃
I have tons on that already, search it on YT yea?
How does the sum of the series come out to less than that of the integral? I always thought that the integral had a higher value, which was why the integral convergence test was valid.
Same question
Why the integral result has smaller value compare the sum result? This result not make sense.
How is that the integral value is smaller than the series value? I thought the integral value should be bigger than the series.
Both are decreasing with increasing x (or n). The sum "holds" the value until the next term, while the integral is decreasing. Just sketch the graphs and you'll see that the step function representing the sum is "on top of" the continuous function.
@@pjmmccann Oh I understand now, Thank you.
@@pjmmccann you can draw the series as a rectangles with width equal to 1 and hight equal to the function but if you start with a rectangle from the left you will see rectangles is less than the continous area so if you substract
from the series [(3/2) - 1/(4-1)]
you get (7/6) which is still bigger than ln(3)
Oscilloscope series
Well well....
2학년의 꿈 ㅇㄷ
Integral 😀
For the first integral, can't I just factor out the 2 as a constant multiple and then end up with 2*[arctan(x)] from 2 to infinity?
Because if you were to go that route, you would get -2 arccoth(x)
You do not even need to limit the 1/(n+1) you dont even need to mention it, because it will be cancelled out by the next-next term. Just as all the other terms except for 1+1/2
That's not quite true. Consider the sum of (n - (n - 1)) from n = 2 to infinity. The terms all cancel except for -1, but the series is 1+1+1+... which obviously shouldn't be -1. The reason it doesn't work in this case is that the leftover term that you're going to cancel later doesn't go to 0 as n goes to infinity.
@@iabervon you are describing a different PROPER sum. This is an IMPROPER telescopic sum, meaning all the terms except for a finite number of them (in the front) eventually cancel out (after the achemedic principle "for every natural number n there is a higher number n+1") . Due to the archimedic principle it only matters what happens with terms of infinite index, not those of finite index.
When you augment "n" 0,5 by 0,5
You will get a value between ln3 and 1,5.
But log3
面白い～♪
Aw, I wanted the answers to be the same haha
Kelfran Gt
Maybe next time! : )
Teacher!! If integral of 1/(X^4+1)dx .
He has done this at least 3 times
@@AviMehra let's go warriors 🏀
You gotta do some partial fractions, completing squares, and substitutions. Also remember derivative of arctan(x) is 1/(x²+1)
@@KelfranGt You can do it? please help me.
Hey to which set does n belong? Like real no. Integers or what?
I see
There are integers I think u got confused
n member of the integers greater than two, or the naturals excluding 1. Meanwhile in the integral, x is a real member of the range [2,inf]
@@AviMehra Nope, you're wrong, because inf is not a real number i.e. in the integral, x is a real member of the range [2, inf)
Bro collar mic is best choice for you rather than that
Well well 😂
Is there a function where the infinite sum and the definite integral from the same number to infinity are equal? No floor function tho, that’s cheating lol
You can graph the sum as 1-wide rectangles whose height matches the curve at the left. If the function is decreasing everywhere, that's obviously bigger.You could probably find a sin(kn+m)/n that worked, though.
I have nothing to say
Yo
Why??
plsss someone explain why 1/n+1 still there ?
You mean 1/(n + 1).
Yow!
First one to comment....black pen red pen
i like infinite series more
The summation result is 3/4 not 3/2.
No? There is a 2 in the numerator. But ye if you go by the thumbnail then you are correct lol.
Seventh
First
fourth
Could we have used a tan substitution for x?
It was my first thought/idea when I've seen this integral LOL. But it's easier solve like him.
However, you have to substitute x by sec(u), not by tan(u) ;-)