# $\sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg]$ [closed]

find the sum of the series, no solution was provided.

$$\sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3} + \sum_{n=0}^\infty \frac{1}{(2n+1)^2}$$

I've tried a few many ways, I'll add my attempts in but they weren't correct

• wolframalpha.com/input/…
– DHMO
Apr 19 '17 at 13:29
• @DMHO yes that works, but it doesn't provide step by step example solution?
– hit
Apr 19 '17 at 13:30
• According to WolframAlpha (link given above), the answer is $\dfrac1{64} (\zeta(3, 1/4) - \zeta(3, 3/4) + 8 \pi^2)≈2.20265$, which is not a nice solution.
– DHMO
Apr 19 '17 at 13:30
• I didn't even finish my comment @hit
– DHMO
Apr 19 '17 at 13:30
• $\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{3}{4}\zeta(2)$ is straightforward; for the remaining part have a look at math.stackexchange.com/questions/850442/… Apr 19 '17 at 16:55

In this mathstackexchange post, an answer shows how a Fourier expansion of $x(1-x)$ on $[0,1]$ gives $$\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}=\dfrac{\pi^3}{32}.$$ On the other hand, we have $$\sum_{n=0}^\infty \frac{1}{(2n+1)^2}=\sum_{n=0}^\infty \frac{1}{n^2}-\sum_{n=0}^\infty \frac{1}{(2n)^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}.$$ So, the two results put together gives $$\sum_{n=0}^\infty \Bigg[ \frac{(-1)^n}{(2n+1)^3} + \frac{1}{(2n+1)^2}\Bigg] =\dfrac{\pi^3}{32}+\dfrac{\pi^2}{8}.$$