The 'Telescope' method seems to be a really clean method of proof, though it requires catching the insight about the successive terms. For the kth term: k.k! = (k+1).k! - k! = (k+1)! - k! As such, each successive value in the series cancels out the first term from the previous value. The only uncancelled values then are the highest (N+1)! and the -1! from the first value. As such: S(n) = (n+1)! - 1 Interestingly, this also works for the '0th' term. S(0) = 0, which makes sense, as (0)0! = 0 The chart method to start off definitely is a good plan, regardless.
Actually when n is large (n > 10), you don't have to remember all these formulae. A very good approximation is as follows : Sum(n) ~= n^2/2 Sum(n^2) ~= n^3/3 Sum(n^3) ~= n^4/4 In General Sum(n^k) ~= n^(k+1)/(k+1) Ring a bell ? It should... Coz the approximation is derived from our famous anti-derivate of power function Integral (x^n) = x^(n+1)/(n+1)
My solution was to just rewrite the "k" in the sum as (k+1)-1, which splits the sum into 2 sums of pure factorials that are slightly offset from each other: Sum((k+1)!) - Sum(k!). Most of the terms cancel out and only the (N+1)! - 1 remains. Interestingly, while the difference of these factorial sums is easy to get, the individual sums seem to be much harder to get an exact formula for. I tried a few methods, including the derivative trick, but I can't seem to get anything very useful.
@@JPiMaths It likely has an official name that I'm not aware of, so it's probably best to show with an example: Suppose you want to evaluate something like S_0 = Sum(k*2^k) where k goes from 1 to N. Instead of attacking the sum directly, you can look at a different sum: S_1 = Sum(x^k), which is a sum of functions. Now, S_1 is just the geometric series, so its result is the well known S_1 = (1-x^(N+1)) / (1-x). Now, suppose we differentiate S_1 with respect to x and evaluate at x = 2. We have an option of differentiating before and after calculating the sum. If we differentiate and evaluate before summing, we get (1/2)*Sum(k*2^k) = (1/2)S_0. On the other hand, if we differentiate after summing, we get 1+(N-1)*2^N. Since our sums are finite, the summation and derivative definitely commute, and these two results must be equal. So, we have S_0 = 2*(1+(N-1)2^N). In general, the trick is applicable whenever we want to sum (polynomial)*(exponential). A similar trick can be done for things like summing (5^(-k))/(2k+1) from 0 to infinity, but then you get an integral instead of a derivative (finite sums of that form are a bit harder as the integral can get complicated, so it isn't as useful). However, in the case of summing factorials, this trick doesn't really help, as we just loop back to the starting point.
One of my favourite things about math is all the different approaches one can take to a problem. I personally started by using the fact that k*k!=k^{2}*(k-1)! (and the fact that 0!=1=1!), set that as my S_{k}, and then used induction to get S_{k+1}=S_{k}+[(k+1)^{2}*k!]. I think your way was more clever. I never would have thought of the S_{k}=(k+1)!-1 equivalence. Very nice proof. EDIT: One of my least favourite things about math is getting caught up in side work and forgetting the plot. I used induction to get each individual term. Ha. Once I actually added all the terms up, I saw your proof better.
@@jonathanevenboer ah yes you ended up just calculating what you originally had calculated. But yes, you'll get better at staying on track with a problem over time.
@@JPiMaths Lol. I'm a grad student in math. I was at the tail end of a night-long session of finding charts to prove a pair of pants is a Riemann Surface. It was brain fatigue more than a beginner's mistake. I left my mistake up (hence the edit and not a deletion) for others who might make the same mistake (or at least a similar mistake).
I solved it in a different way. S = Σk(k!) = Σ(k+1-1)(k!) = Σ(k+1)(k!) - Σ(k!)= Σ(k+1)! - Σ(k!) = 2!-1!+3!-2!+4!-3!..... Starting from 2!, all terms get cut off except -1! and (n+1)!. So, answer is S = (n+1)! - 1
I tried solving the question before watching the video. I solved it a different way. After playing around, I noticed that Σ(k+1)! = Σ(k+1)k! = Σk(k!) + Σk! Or rewritten, Σk(k!) = Σ(k+1)! -Σk!. But the RHS is simply a telescoping sum, which gives the result you got after doing the method of differences Thank you for such a good question!
Your kidding with us. This problem is so easy that it didn't take 10 seconds of my time. This wouldn't be asked in a hard exam. This question is too easy. Btw the answer is (n+1)!-1
@@CallMeIcebrick yeah I know😂. I guess some questions are solvable and some are very hard. There is no easy questions. Even at the easiest ones you should think for a while. So this question may be asked. Because it takes even a little while to answer
The 'Telescope' method seems to be a really clean method of proof, though it requires catching the insight about the successive terms.
For the kth term:
k.k! = (k+1).k! - k! = (k+1)! - k!
As such, each successive value in the series cancels out the first term from the previous value. The only uncancelled values then are the highest (N+1)! and the -1! from the first value. As such:
S(n) = (n+1)! - 1
Interestingly, this also works for the '0th' term. S(0) = 0, which makes sense, as (0)0! = 0
The chart method to start off definitely is a good plan, regardless.
@@HeavyMetalMouse telescoping is good!!
Actually when n is large (n > 10), you don't have to remember all these formulae. A very good approximation is as follows :
Sum(n) ~= n^2/2
Sum(n^2) ~= n^3/3
Sum(n^3) ~= n^4/4
In General Sum(n^k) ~= n^(k+1)/(k+1)
Ring a bell ? It should... Coz the approximation is derived from our famous anti-derivate of power function Integral (x^n) = x^(n+1)/(n+1)
@@agytjax ok that's quite cool, I'd not seen than before
@@JPiMaths - Glad it was useful 😀
My solution was to just rewrite the "k" in the sum as (k+1)-1, which splits the sum into 2 sums of pure factorials that are slightly offset from each other: Sum((k+1)!) - Sum(k!). Most of the terms cancel out and only the (N+1)! - 1 remains.
Interestingly, while the difference of these factorial sums is easy to get, the individual sums seem to be much harder to get an exact formula for. I tried a few methods, including the derivative trick, but I can't seem to get anything very useful.
@@kriegsmesser4567 what do you mean by the derivative trick?
@@JPiMaths It likely has an official name that I'm not aware of, so it's probably best to show with an example:
Suppose you want to evaluate something like S_0 = Sum(k*2^k) where k goes from 1 to N. Instead of attacking the sum directly, you can look at a different sum: S_1 = Sum(x^k), which is a sum of functions. Now, S_1 is just the geometric series, so its result is the well known S_1 = (1-x^(N+1)) / (1-x). Now, suppose we differentiate S_1 with respect to x and evaluate at x = 2. We have an option of differentiating before and after calculating the sum. If we differentiate and evaluate before summing, we get (1/2)*Sum(k*2^k) = (1/2)S_0. On the other hand, if we differentiate after summing, we get 1+(N-1)*2^N. Since our sums are finite, the summation and derivative definitely commute, and these two results must be equal. So, we have S_0 = 2*(1+(N-1)2^N).
In general, the trick is applicable whenever we want to sum (polynomial)*(exponential). A similar trick can be done for things like summing (5^(-k))/(2k+1) from 0 to infinity, but then you get an integral instead of a derivative (finite sums of that form are a bit harder as the integral can get complicated, so it isn't as useful). However, in the case of summing factorials, this trick doesn't really help, as we just loop back to the starting point.
@@kriegsmesser4567 Ah yes, thanks for explaining. I am aware of the trick, but I also don't think it has an official name
I got asked a problem very similar to one in the TBO booklet in my interview today! Thank you ❤
@Theproofistrivial ah nice! Hoping the interview went well! Which problem was it?
One of my favourite things about math is all the different approaches one can take to a problem.
I personally started by using the fact that k*k!=k^{2}*(k-1)! (and the fact that 0!=1=1!), set that as my S_{k}, and then used induction to get S_{k+1}=S_{k}+[(k+1)^{2}*k!].
I think your way was more clever. I never would have thought of the S_{k}=(k+1)!-1 equivalence. Very nice proof.
EDIT: One of my least favourite things about math is getting caught up in side work and forgetting the plot. I used induction to get each individual term. Ha. Once I actually added all the terms up, I saw your proof better.
@@jonathanevenboer ah yes you ended up just calculating what you originally had calculated. But yes, you'll get better at staying on track with a problem over time.
@@JPiMaths Lol. I'm a grad student in math. I was at the tail end of a night-long session of finding charts to prove a pair of pants is a Riemann Surface. It was brain fatigue more than a beginner's mistake. I left my mistake up (hence the edit and not a deletion) for others who might make the same mistake (or at least a similar mistake).
I just added sum of n! and then it becomes a sum up to (n+1)! and it goes from there
k*k!
=(k+1-1)*k!
=(k+1)*k!-k!
=(k+1)!-k!
The telescoping sum equals (n+1)!-1.
I solved it in a different way.
S = Σk(k!) = Σ(k+1-1)(k!) = Σ(k+1)(k!) - Σ(k!)= Σ(k+1)! - Σ(k!) = 2!-1!+3!-2!+4!-3!.....
Starting from 2!, all terms get cut off except -1! and (n+1)!.
So, answer is S = (n+1)! - 1
@pradhyunmudaliar6606 this is a nice solve too!
@@JPiMaths Thanks!
it can be easily proved using induction
@@sudoku_891 yes, I think that would make for a good video
You put Worcester college in the thumbnail, is that the college you attended?
@Theproofistrivial I was at Christ Church!
I tried solving the question before watching the video. I solved it a different way.
After playing around, I noticed that Σ(k+1)! = Σ(k+1)k! = Σk(k!) + Σk!
Or rewritten, Σk(k!) = Σ(k+1)! -Σk!. But the RHS is simply a telescoping sum, which gives the result you got after doing the method of differences
Thank you for such a good question!
@@nukeeverything1802 bingo! Nice spot!!
So basically you solve it by knowing the answer from the book. Nice! :)
No, you just poke and prod the relationship and then yes proof by induction. If you do enough proofs it becomes second nature
Hey I like your channel, nice problems. I like looking at you because you remind me of Ramanujan😭😂
@alphazero339 haha thank you. What exactly about me reminds you of Ramanujan? 😅
@JPiMaths I'm not sure maybe you seem smart and solve nice series and from pictures of him I saw its also little similar XD
No. That’s why I studied english there instead
@@rabbitrun777 😂😂😂
trop facile
Your kidding with us. This problem is so easy that it didn't take 10 seconds of my time. This wouldn't be asked in a hard exam. This question is too easy.
Btw the answer is (n+1)!-1
You smell
@@CallMeIcebrick yeah I know😂. I guess some questions are solvable and some are very hard. There is no easy questions. Even at the easiest ones you should think for a while. So this question may be asked. Because it takes even a little while to answer