Question

So I stumped here:
https://brilliant.org/problems/inspired-by-calvin-lin-3/?group=9gTI39kmvZXz

I figured that this series equals to \(\displaystyle \sum_{n=1}^\infty\left[\sum_{i=1}^\infty\dfrac{1}{(4n)^{2i}}\right]\) and I managed to figure that inner series equals to \(\dfrac{1}{16n^2-1}\)

So here's where I am stuck at;

\[\Large\sum_{n=1}^\infty\dfrac{1}{16n^2-1}\]How in the world can I evaluate this?

Answer 1

@igreen @hero @uri @preetha

Answer 2

wolfram says that evaluates to \(\large \frac{4-\pi}{8}\)

http://www.wolframalpha.com/input/?s=53&_=1430886163540&fp=1&i=%5csum%5climits_%7bn%3d1%7d%5e%7b%5cinfty%7dzeta(2n)%2f4%5e(2n)&incTime=true

Answer 3

Yeah, but I'm looking for a way to solve it by hand.

Answer 4

Not sure if this is possible though

Answer 5

\[\frac{1}{16n^2-1}=\frac{A}{4n-1}+\frac{B}{4n+1}\]
guess this won't help probably

Answer 6

telescoping is out because the final answer is not a rational number

Answer 7

@phi @dan815 @wio @zepdrix

Answer 8

@amistre64

Answer 9

Does this help? http://math.ucsb.edu/~cmart07/Evaluating%20Series.pdf

Answer 10

Not really, or at least I cannot see how to apply few of these methods.

Answer 11

I am thinking on Wallis's Product

Answer 12

\[\sum_1\frac1{(16n^2)^{i}}\]
\[\frac{1}{1-\frac{1}{16n^2}}-1\]

\[\frac{16n^2}{16n^2-1}-1\]
\[\frac{1}{16n^2-1}=\frac{1}{8n-2}-\frac{1}{8n+2}\]

i broke the wolf ... the left side converges, but it told me the right side does not :)

Answer 13

There are many similarities in this which I assume there must be a connection on how to get the answer: \[\frac{ 4-\Pi }{ 8 }\] -------------------------\[\frac{ 2 }{ \Pi } = \prod_{n=1}^{\infty} (1-\frac{ 1 }{ 4n^2 })\]\[\frac{ \Pi }{ 2 } = \prod_{n=1}^{\infty} (\frac{ 4n^2 }{ 4n^2 -1 })\]\[\frac{ \Pi }{ 2 } = \prod_{n=1}^{\infty} (\frac{ (2n)(2n) }{ (2n-1)(2n+1) })\]