I'm trying to solve the following problem:
Find the sum of the series $\sum^{\infty}_{n=1}\frac{(-1)^n}{(2n+1)^3}$ by using the function $f(z)=|(2z+1)^3\sin\pi z|^{-1}.$
I can't see it... I do see that if I define $g(z)=\frac{1}{(2n+1)^3\sin\pi z},$ then I have $ f(z)=|g(z)| $ and $ \sum^{\infty}_{n=1}\frac{(-1)^n}{(2n+1)^3}=\sum^{\infty}_{n=1}g(n). $
[NOT TRUE, actually -- added a day later.]
But I can't see why it's good do look at the absolute value of $g$ instead of just $g,$ and I just don't know how looking at the continuous function on all $\mathbb C$ can help me with this discrete sum...
Could you please give me some hints? The problem should probably be easy so I may just need a slight push, but I'm not sure of course.
Re comments. Thank you. I've calculated the poles of the function $g$ and their residues. (I don't know how I could do it for $f$. I think it's not a holomorphic function because it has only real values.) Here are my calculations: $\begin{eqnarray} (2z+1)^3\sin\pi z=0\\ z=-\frac{1}{2}\vee z\in \mathbb Z \end{eqnarray} $ So these are the poles. The residue at $-\frac{1}{2}$ is $-\frac{1}{8}$ because $\lim_{z\to -\frac{1}{2}}\frac{(z+\frac{1}{2})^3}{(2z+1)^3\sin\pi z}=\frac{1}{8\sin(-\frac{\pi}{2})}=-\frac{1}{8}. $
The residues at the integers $n\in\mathbb Z$ I calculated this way:
$\begin{eqnarray} \lim_{z\to n}\frac{z-n }{(2z+1)^3\sin\pi z } &=& \lim_{x\to 0}\frac{x}{(2(x+n)+1)^3\sin\pi(x+k) }\\ &=& \lim_{x\to 0} \frac{(-1)^n x}{(2x+2n+1)^3\sin\pi x}\\ &=& \lim_{x\to 0} \frac{(-1)^n \pi x}{\pi (2x+2n+1)^3\sin\pi x}\\ &=& \frac{(-1)^n}{\pi (2n+1)^3} \end{eqnarray} $
So I have $ \sum_{n=1}^{\infty}\frac{(-1)^n}{(2n+1)^3}=\pi\sum_{n=1}^\infty \mathrm{Res}_n g. $
Now I don't know how the sum of all residues of the positive poles of $g$ is connected to the function $|g|,$ which I am supposed to use...
Some new thoughts I haven't given up on this problem even though I still don't see how to solve it. I hope it's because I did too little myself that there have been only so laconical responses so far and perhaps after this edit someone will give me the help I desperately need.
I already know what the residues are of the function $g$ (if I calculated them correctly). I know that I need to find the sum of all residues at the positive poles. This would be the limit (modulo a scalar multiplication) of the sequence of line integrals along curves encircling each of more and more of the positive poles of $g.$ Perhaps I could add some more line integrals to such a sequence and prove by some estimation that it diverges to zero.
I would need to be able to estimate the absolute value of $g$ on those curves. (This gives me hope that I'm on the right track, because $f=|g|$ was mentioned in the problem.) I think it might be a good idea to use the following estimation:
$\mbox{intergal}\leq (\mbox{upper bound of }|g|\mbox{ on }\gamma_n)\cdot (\mbox{length of curve}).$
The length of the curve will probably be linear with respect to $n$ so I will need the bound to be at most of the order $n^{-2}$ to prove that the integrals go to zero with $n\to\infty.$
But I don't know how to choose the sequence of curves and how to estimate the absolute value of $\sin\pi z...$
I would be very grateful if someone could tell me if my approach is right. And if it is for guidance in the specifics. In particular, I have no idea how I could try to bound $|\sin\pi z|.$ I know that there is no global scalar bound but I believe a linear bound could be enough.