Prove the series:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$
Solution:
Successively we have:
$$\begin{aligned} 1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\ &\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\ &=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\ &= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )\\
&=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\ &=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\ &=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32} \end{aligned}$$
$(*)$ is justified due to absolute converegence.
Update: A solution with complex analysis could be the following:
We are invoking the kernel function \( \pi \csc \pi z \). The function \( \displaystyle g(z)=\frac{1}{\left ( 2z-1 \right )^3} \) has clearly one pole of order \(3 \) at \(z=1/2 \). Let \( N \) be a positive integer and consider the contour integral:
$$\oint_{\Gamma_N}\frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}\, {\rm d}z$$
where \( \Gamma_N \) is a positive oriented square with vertices \( \displaystyle \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ) \). Then by the residue thoerem we have:
$$\frac{1}{2\pi i }\oint_{\Gamma_N}\frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}\,{\rm d}z=\sum_{n=-N}^{N}\mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}+\mathfrak{Res}_{z=\frac{1}{2}}\frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}$$
Letting \( N \rightarrow +\infty \) the contour integral goes to \( 0 \) since for $$ |z| \geq 1 \implies \left | \frac{1}{\left ( 2z-1 \right )^3} \right |\leq \frac{1}{|z|^3} $$ hence:
$$0=\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}}{\left ( 2n-1 \right )^3}+\frac{\pi^3}{16}\Leftrightarrow \sum_{n=-\infty}^{\infty}\frac{(-1)^{n-1}}{\left ( 2n-1 \right )^3}=\frac{\pi^3}{16}$$
However:
$$\begin{aligned}\sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3} &= \sum_{n = -\infty}^0 \frac{(-1)^{n-1}}{(2n-1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n+1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.
\end{aligned}$$
Therefore:
$$\frac{\pi^3}{16} = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}$$
and we are done.
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^3}=\frac{\pi^3}{32}$$
Solution:
Successively we have:
$$\begin{aligned} 1-\frac{1}{3^3}+\frac{1}{5^3}-\cdots &=\sum_{n=0}^{\infty}\frac{(-1)^n}{\left ( 2n+1 \right )^3} \\ &\overset{(*)}{=} \left ( 1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots \right )-\left ( \frac{1}{3^3}+\frac{1}{7^3}+\frac{1}{11^3}+\cdots \right )\\ &=\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+1 \right )^3} \; -\sum_{n=0}^{\infty}\frac{1}{\left ( 4n+3 \right )^3} \\ &= -\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{1}{4} \right )+\frac{1}{2\cdot 4^3}\psi^{(2)}\left ( \frac{3}{4} \right )\\
&=\frac{1}{2\cdot 4^3}\left [ \psi^{(2)}\left ( 1-\frac{1}{4} \right )-\psi^{(2)}\left ( \frac{1}{4} \right ) \right ]\\ &=\frac{1}{2\cdot 4^3}\left [ 2\pi^3 \cot \frac{\pi}{4} \csc^2 \frac{\pi}{4} \right ] \\ &=\frac{\pi^3 \cot \frac{\pi}{4}\csc^2 \frac{\pi}{4}}{4^3}=\frac{\pi^3}{32} \end{aligned}$$
$(*)$ is justified due to absolute converegence.
Update: A solution with complex analysis could be the following:
We are invoking the kernel function \( \pi \csc \pi z \). The function \( \displaystyle g(z)=\frac{1}{\left ( 2z-1 \right )^3} \) has clearly one pole of order \(3 \) at \(z=1/2 \). Let \( N \) be a positive integer and consider the contour integral:
$$\oint_{\Gamma_N}\frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}\, {\rm d}z$$
where \( \Gamma_N \) is a positive oriented square with vertices \( \displaystyle \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ) \). Then by the residue thoerem we have:
$$\frac{1}{2\pi i }\oint_{\Gamma_N}\frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}\,{\rm d}z=\sum_{n=-N}^{N}\mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}+\mathfrak{Res}_{z=\frac{1}{2}}\frac{\pi \csc \pi z}{\left ( 2z-1 \right )^3}$$
Letting \( N \rightarrow +\infty \) the contour integral goes to \( 0 \) since for $$ |z| \geq 1 \implies \left | \frac{1}{\left ( 2z-1 \right )^3} \right |\leq \frac{1}{|z|^3} $$ hence:
$$0=\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}}{\left ( 2n-1 \right )^3}+\frac{\pi^3}{16}\Leftrightarrow \sum_{n=-\infty}^{\infty}\frac{(-1)^{n-1}}{\left ( 2n-1 \right )^3}=\frac{\pi^3}{16}$$
However:
$$\begin{aligned}\sum_{n = -\infty}^\infty \frac{(-1)^{n-1}}{(2n-1)^3} &= \sum_{n = -\infty}^0 \frac{(-1)^{n-1}}{(2n-1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = \sum_{n = 0}^\infty \frac{(-1)^{n}}{(2n+1)^3} + \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}\\
& = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}.
\end{aligned}$$
Therefore:
$$\frac{\pi^3}{16} = 2\sum_{n = 0}^\infty \frac{(-1)^n}{(2n+1)^3}$$
and we are done.
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