# The sum of a series to be found by looking at a complex function.

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.

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Are you doing things with the residue theorem perhaps? – mixedmath Jan 30 '12 at 1:28
@mixedmath I'd be glad to start doing them! I've read some basic theorems. But I still don't see the connection... – Bartek Jan 30 '12 at 1:31
@Bartek, analyze what your function $f$ "looks" like in the complex plane. In particular, what is the singularity structure? Where are the poles located, and what residues do they have? Once you answer these questions I'm sure you'll start thinking in the right direction. – Ragib Zaman Jan 30 '12 at 9:51
@RagibZaman I've edited my post. How can I calculate the residues of $f$? It's not a holomorphic function, right? – Bartek Jan 30 '12 at 13:35

Consider the following contour integral taken around bigger and bigger circles around the origin until it circles the whole complex plane: $$\oint\frac{\pi\csc(\pi z)}{(2z+1)^3}\mathrm{d}z=0\tag{1}$$ This is so because $\csc(\pi z)$ vanishes exponentially off of $\mathbb{R}$, and the integral passing through the real line at half-integer points gets smaller as the radius gets bigger.
Note that $\pi\csc(\pi z)$ has poles with residue $(-1)^n$ at every integer $n$. Equation $(1)$ says that the sum of all the residues of $\dfrac{\pi\csc(\pi z)}{(2z+1)^3}$ is $0$; that is, $$\sum_{n=-\infty}^\infty\frac{(-1)^n}{(2n+1)^3}+\operatorname{Res}\left(\frac{\pi\csc(\pi z)}{(2z+1)^3},-\frac12\right)\tag{2}$$ Note that translating $z\mapsto z-\frac12$ yields \begin{align} \operatorname{Res}\left(\frac{\pi\csc(\pi z)}{(2z+1)^3},-\frac12\right) &=\operatorname{Res}\left(-\frac{\pi\sec(\pi z)}{8z^3},0\right)\\ &=-\frac{\pi}{8}\operatorname{Res}\left(\frac{1+\frac12\pi^2z^2+\dots}{z^3}\right)\\ &=-\frac{\pi^3}{16}\tag{3} \end{align} Combining $(2)$ and $(3)$ and dividing by $2$ gives $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}\tag{4}$$ Subtracting $1$ from both sides of $(4)$ yields $$\sum_{n=1}^\infty\frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}-1\tag{5}$$
Thank you very much! I tried to do it exactly the way you've done it but it didn't work. I couldn't see why (4) should be true. But I thought about it again, now that I knew it was the correct way, and I see that there is a symmetry: $\frac{(-1)^{-n-1}}{(2(-n-1)+1)^3}=\frac{(-1)^n}{(2n+1)^3}.$ Great! – Bartek Jan 31 '12 at 2:41