Question

how to get this sum:
\[\sum_{1}^{\infty}(-1)^{n+1}\frac{1}{n^5}\]
I know it converges by the alternating series test
but can't find a way to for the sum

Answer 1

hmm wolfram mentioned something about the riemann zeta function
you know anything about that function?

Answer 2

yeah i checked the wolfram, but we didn't do any do at all

Answer 3

of that *

Answer 4

reimann zeta of 5 is computed with integral yes

Answer 5

\[\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n^5}=\left(\sum_{n=1}^{\infty}\frac{1}{(2n-1)^5}\right)-\left(\sum_{n=1}^{\infty}\frac{1}{(2n)^5}\right)\]Yes?

Answer 6

hmm didn't get where did you got that lol

Answer 7

For every odd n, \((-1)^{n+1} = 1\) and for every even n, \((-1)^{n+1} = -1\), right?

So what I did was arrange these terms so that positive and negative terms are grouped separately.

Answer 8

After you said I was thinking telescoping series but that looks pretty ugly to me :(

Answer 9

is that what you were going for ?

Answer 10

Hmm I have no idea what to do next XD

I was basically suggested to arrange terms to perhaps make this problem easier.

Answer 11

well you did more than I could
I'm kinda like drawing a blank

Answer 12

just throw some thoughts. That's all.

Answer 13

hmm i see,
sorry had some lag here lol

Answer 14

that one is wrong!

Answer 15

I know...

Answer 16

I somehow see \(5^n\), you know?

Answer 17

But you have any idea on \[\sum_{n=1}^{\infty}\frac{1}{(2n-1)^5}\] part?

Answer 18

i did see it that way first time lol

Answer 19

that still not easy to deal with lol

Answer 20

that's not a telescoping series is it?

Answer 21

@ganeshie8 see if you can help :)

Answer 22

I doubt it.

Answer 23

About telescoping serie lol

Answer 24

hmm integral test would work with that one

Answer 25

\[\begin{align}\sum_{1}^{\infty}(-1)^{n+1}\frac{1}{n^5}
&=\sum_{1}^{\infty}\frac{1}{(2n-1)^5} - \frac{1}{(2n)^5}\\~\\
&=\sum_{1}^{\infty}\frac{1}{(2n-1)^5}\color{blue}{+\frac{1}{(2n)^5}-\frac{1}{(2n)^5}} - \frac{1}{(2n)^5}\\~\\
&=\sum_{1}^{\infty}\frac{1}{n^5} - \frac{2}{(2n)^5}\\~\\
&=\sum_{1}^{\infty}\frac{1}{n^5} - \frac{1}{16n^5}\\~\\
&=\frac{15}{16}\sum_{1}^{\infty}\frac{1}{n^5}\\~\\
&=\frac{15}{16}\zeta(5)
\end{align}\]

Answer 26

Smart move.