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Post Tue Nov 22, 2016 5:32 pm

Posts: 82
Calculation of $\displaystyle\sum^{\infty}_{n=0}\frac{(-1)^n}{(2n+1)^3}$

Post Fri Nov 25, 2016 11:57 pm

Posts: 10
First we note that

$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{(2n+1)^3}=2S$$

where $S$ is your sum.

Then we may apply the following residue formula*:

$$\sum_{n=-\infty}^{\infty}(-1)^n\frac{P(n)}{Q(n)}=-\pi\sum_{\sigma\in K}\underset{z=\sigma}{Res}\left [ \frac{P(z)}{Q(z)\sin(\pi z)}\right ]$$

where $K$ is the set of locations of poles of $P(z)/Q(z)$ in $\mathbb{C}$.

Here we have $P(z)\overset{def}{=}1$, $Q(z)=(2z+1)^3$, and $K=\left\{-1/2\right\}$ so

$$\begin{align*}
2S &= -\pi\underset{z=-1/2}{Res}\left [ \frac{1}{(2z+1)^3\sin(\pi z)}\right ] \\
&= \frac{\pi^3}{16}.
\end{align*}$$

This implies that $S=\frac{\pi^3}{32}$.

*Note that this formula works only when $Q(z)=0$ has no integer solutions and we must have $degree(Q(z))-degree(P(z))\geq2$.
The internet is a $\sum^{\infty}_{n=0}$tube$_n$ filled with cats!



Post Sun Nov 27, 2016 2:19 pm

Posts: 82
Thanks kenesin


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