BEFORE you can scramble the terms, you need to know this series converges absolutely. As you allude to but don’t use explicitly, you can compare this series to the geometric series with general term 1/(2n)^2 and easily determine the absolute convergence, making the partial sums and term scramble a clever solution to getting the sum.
Can anyone give an example where this is actually a problem and leads you astray? I always see people saying this but it always seems to work out perfectly anyway.
You have to be careful with that sum of the split fractions, as its positive terms on their own diverge to +∞, and its negative terms on their own diverge to -∞, and the series as a whole converges, so if you re-order the terms you can get different values. (The Riemann rearrangement theorem) I don't know about re-arranging the parentheses like you did though; I assume that is safe?
@@mbmillermo No that's something else: a generalised form ("analytic continuation") of the Riemann zeta function. For the Riemann rearrangement theorem, you need the positive and negative terms to diverge when separated, but the series as a whole to converge.
Can anyone give an example where this is actually a problem and leads you astray? I always see people saying this but it always seems to work out perfectly anyway.
It is totally safe. Besides, the parentheses do not need to be removed in the infinite sum. They only need to be removed when evaluating the finite partial sums, and in a finite sum, parentheses can always be removed.
It seems like a fairly standard exercise on telescopic series, once we learn the trick of the Mengoli series it always leads us back to the decomposition with partial fractions.
Once you say what the sum is, the whole thing is a well-known telescoping sum. However, from what is given (1/3+1/15+1/35+...) it is impossible to guess what sum you have in mind.
I came up with the approximation "more than pi^2/24", because your sum is 1/(4*1^2-1) + 1/(4*2^2-1) + 1/(4*3^2-1) + 1/(4*4^2-1) + ..., while the Riemann zeta function, evaluated at s=2, and divided by four, is 1/(4*1^2) + 1/(4*2^2) + 1/(4*3^2) + 1/(4*4^2) + .... Since the denominators in your series are slightly smaller, I surmised your sum would be slightly larger. I admit that I am surprised that it's nearly 22% larger, though.
The title is misleading. It is an infinite series but the sum is finite. You can do that indeed with partial fraction decomposition OR with powerseries...
If the series converges to (1/2), then why do you title the video “This is not a finite sum?” The number of terms of the series may not be finite. The sequence of partial sums generated by successive terms of the series may go on forever. That sequence ultimately converges to one-half, though. Good video otherwise.
If you start with n = 0 (and go to n = N letting N go towards infinity) you will get that the sum is S = (1/2)[1/(2n + 1) - 1/(2n + 3)] = (1/2)[1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... - 1/(2N + 1) + 1/(2N + 1) - 1/(2N + 3)] = lim(1/2)*[1 - 1/(2N + 3)] = 1/2*[1] = 1/2.
BEFORE you can scramble the terms, you need to know this series converges absolutely. As you allude to but don’t use explicitly, you can compare this series to the geometric series with general term 1/(2n)^2 and easily determine the absolute convergence, making the partial sums and term scramble a clever solution to getting the sum.
a GP series has a varying exponent say 'n'.The first term and common ratio are constant...your general term is wrong!
@@mrityunjaykumar4202 Thx. I meant to say the p-series, not the geometric series. No excuse.
Can anyone give an example where this is actually a problem and leads you astray? I always see people saying this but it always seems to work out perfectly anyway.
You have to be careful with that sum of the split fractions, as its positive terms on their own diverge to +∞, and its negative terms on their own diverge to -∞, and the series as a whole converges, so if you re-order the terms you can get different values. (The Riemann rearrangement theorem)
I don't know about re-arranging the parentheses like you did though; I assume that is safe?
Is that the thing that gives us 1+2+3+...+∞ = -1/12?
@@mbmillermo No that's something else: a generalised form ("analytic continuation") of the Riemann zeta function.
For the Riemann rearrangement theorem, you need the positive and negative terms to diverge when separated, but the series as a whole to converge.
Can anyone give an example where this is actually a problem and leads you astray? I always see people saying this but it always seems to work out perfectly anyway.
It is totally safe. Besides, the parentheses do not need to be removed in the infinite sum. They only need to be removed when evaluating the finite partial sums, and in a finite sum, parentheses can always be removed.
That was beautiful! So deceivingly simple!
Thank you! 😊
Learned a very cool technique. Thank you.
Glad it was helpful!
Simply amazing! Wow 😍
Very nice! Thank you.
Thank you too!
It seems like a fairly standard exercise on telescopic series, once we learn the trick of the Mengoli series it always leads us back to the decomposition with partial fractions.
I've never used partial fractions like this 😮great vid!!
Glad it was helpful!
Beautiful question... learn alot...❤❤❤.
Glad you liked it
"You have to be careful when you rearrange the"
No you don't 😎😎😎 I'm never careful and it works every time
That's awesome! 🤩
Yes.
Very good!
Thanks!
Cool!
Once you say what the sum is, the whole thing is a well-known telescoping sum. However, from what is given (1/3+1/15+1/35+...) it is impossible to guess what sum you have in mind.
I failed exactly this question yesterday on the exam smh
I came up with the approximation "more than pi^2/24", because your sum is 1/(4*1^2-1) + 1/(4*2^2-1) + 1/(4*3^2-1) + 1/(4*4^2-1) + ..., while the Riemann zeta function, evaluated at s=2, and divided by four, is 1/(4*1^2) + 1/(4*2^2) + 1/(4*3^2) + 1/(4*4^2) + .... Since the denominators in your series are slightly smaller, I surmised your sum would be slightly larger. I admit that I am surprised that it's nearly 22% larger, though.
👍
Nice! We call this trik telescopic 😂
Yes 😁
I noticed that your terms are also ¼*(1/[n²-¼]) > ¼*(1/n²), so the answer has to be more than π²/24 which is approximately 0.411.
nice :)
Thanks!
easy
*@ SyberMath* -- It is an infinite series, but it has a *finite* sum of 1/2. You need to change the title.
I changed the title to reflect that 🤩
1/2
The title is misleading. It is an infinite series but the sum is finite. You can do that indeed with partial fraction decomposition OR with powerseries...
I changed the title to reflect that 🤩
If the series converges to (1/2), then why do you title the video “This is not a finite sum?” The number of terms of the series may not be finite. The sequence of partial sums generated by successive terms of the series may go on forever. That sequence ultimately converges to one-half, though.
Good video otherwise.
Convergent does not mean finite.
Thanks!
I changed the title to reflect that 🤩
An infinite series with a finite sum converges. Otherwise, it diverges.
I changed the title to reflect that 🤩
X=1/2
If you start with n = 0 (and go to n = N letting N go towards infinity) you will get that the sum is S = (1/2)[1/(2n + 1) - 1/(2n + 3)] = (1/2)[1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... - 1/(2N + 1) + 1/(2N + 1) - 1/(2N + 3)] = lim(1/2)*[1 - 1/(2N + 3)] = 1/2*[1] = 1/2.