${\sum _{n=0}^{\infty }\left(\sum _{i=n+1}^{\infty } \frac{(-1)^{i+1}}{i}\right) \frac{1}{2 n+1} = \frac{\pi^2}{16}}$; is this known?

Question

I was playing around with some series and integrals and I found an interesting relationship. Throughout, let $n\in\mathbb{N}^+$ and let $H_{\alpha}$ be the generalized harmonic number, $\displaystyle{H_{\alpha} = \int _{0}^{1} \frac{1-x^{\alpha}}{1-x}\,dx}$; below, the $4$ is just to make the arithmetic nicer. We have $$ 4 \int _{0}^{\pi/4} \tan^{2n+1}(x)\,dx = H_{n/2}-H_{(n-1)/2} $$We can't just add up in $n$ as the harmonic series diverges. However, if we weight against $(-1)^{n}/(2n+1)$, the series will converge by Dirichlet's Test [also interesting is weighting against just $(-1)^n$], and it simplifies to $$ 4\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int _{0}^{\pi/4} \tan^{2n+1}(x)\,dx = 4 \int _{0}^{\pi/4}\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} (\tan(x))^{2n+1}\,dx $$ $$ =4 \int _{0}^{\pi/4}\arctan(\tan(x))\,dx = 4 \int _{0}^{\pi/4} x\,dx = \frac{\pi^2}{8} $$Using the duplication identities, we have $$ H_{n/2}-H_{(n-1)/2} = 2 \left(\log(2)-\sum_{i=1}^{n}\frac{(-1)^{i+1}}{i}\right) = 2\sum_{i=n+1}^{\infty}\frac{(-1)^{i+1}}{i} $$So together this gives $$ \sum _{n=0}^{\infty }\left(\sum _{i=n+1}^{\infty } \frac{(-1)^{i+1}}{i}\right) \frac{1}{2 n+1} = \frac{\pi^2}{16} $$A related sum is $$ \sum _{n=0}^{\infty } (-1)^n \left(H_{n/2}-H_{(n-1)/2}\right)=1 $$Are these known? Mathematica doesn't seem to recognize either.

Answer 1

Let $ n $ be a positive integer.

We have the following : \begin{aligned}\left|\sum_{k=1}^{n}{\frac{\left(-1\right)^{k}}{2k+1}\int_{0}^{1}{\frac{x^{2k+1}}{1+x^{2}}\,\mathrm{d}x}}-\int_{0}^{1}{\frac{\arctan{x}}{1+x^{2}}\,\mathrm{d}x}\right|&=\left|\int_{0}^{1}{\frac{1}{1+x^{2}}\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{x^{2k+1}}{2k+1}}\,\mathrm{d}x}\right|\\ &\leq\int_{0}^{1}{\frac{1}{1+x^{2}}\left|\sum_{k=n+1}^{+\infty}{\left(-1\right)^{k}\frac{x^{2k+1}}{2k+1}}\right|\mathrm{d}x}\\ &\leq\frac{1}{2n+3}\int_{0}^{1}{\frac{x^{2n+3}}{1+x^{2}}\,\mathrm{d}x}\\ &\leq\frac{1}{2n+3}\int_{0}^{1}{\frac{\mathrm{d}x}{1+x^{2}}}\underset{n\to +\infty}{\longrightarrow}0\end{aligned}

Thus : $$ \sum_{n=1}^{+\infty}{\frac{\left(-1\right)^{n}}{2n+1}\int_{0}^{1}{\frac{x^{2n+1}}{1+x^{2}}\,\mathrm{d}x}}=\lim_{n\to +\infty}{\sum_{k=1}^{n}{\frac{\left(-1\right)^{k}}{2k+1}\int_{0}^{1}{\frac{x^{2k+1}}{1+x^{2}}\,\mathrm{d}x}}}=\int_{0}^{1}{\frac{\arctan{x}}{1+x^{2}}\,\mathrm{d}x} $$

Substituting $ \small\left\lbrace\begin{aligned}y&=x^{2}\\ \mathrm{d}y&=2x\,\mathrm{d}x\end{aligned}\right. $, we get that $ \left(-1\right)^{n}\int_{0}^{1}{\frac{x^{2n+1}}{1+x^{2}}\,\mathrm{d}x}=\frac{\left(-1\right)^{n}}{2}\int_{0}^{1}{\frac{x^{n}}{1+x}\,\mathrm{d}x}=\frac{1}{2}\sum\limits_{k=n+1}^{+\infty}{\frac{\left(-1\right)^{k-1}}{k}} $, thus : \begin{aligned} \sum_{n=1}^{+\infty}{\frac{1}{2n+1}\sum_{k=n+1}^{+\infty}{\frac{\left(-1\right)^{k-1}}{k}}}&=2\int_{0}^{1}{\frac{\arctan{x}}{1+x^{2}}\,\mathrm{d}x}\\ &=\left[\arctan^{2}{x}\right]_{0}^{1}\\ &=\frac{\pi^{2}}{16} \end{aligned}

Answer 2

I can't find any reference to them, but one can solve this problem using series manipulations and generating functions. We being by noting that we are summing over all values of $i>n$, and so we can reindex to get that

\begin{align*} S&=\sum_{n=0}^{\infty}\left(\sum_{i=n+1}^{\infty}\frac{\left(-1\right)^{i+1}}{i}\right)\frac{1}{2n+1}\\ &=\sum_{i=1}^{\infty}\left(\sum_{n=0}^{i-1}\frac{1}{2n+1}\right)\frac{\left(-1\right)^{i+}}{i}\\ &=\sum_{i=1}^{\infty}\left(H_{2i-1}-\frac{1}{2}H_{i-1}\right)\frac{\left(-1\right)^{i+1}}{i}\\ &=\sum_{i=1}^{\infty}H_{2i-1}\frac{\left(-1\right)^{i+1}}{i}-\frac{1}{2}\sum_{i=1}^{\infty}H_{i-1}\frac{\left(-1\right)^{i+1}}{i} \end{align*}

We can now recall the well known generating function

$$\sum_{n=1}^{\infty}H_nx^n=-\frac{\ln(1-x)}{1-x}$$

Which we can easily rearrange and integrate to get our two sums. Namely,

$$\sum_{n=1}^{\infty}H_{n-1}\frac{\left(-1\right)^{n+1}}{n}=-\frac{1}{2}\ln^{2}\left(2\right)$$

and

$$\sum_{n=1}^{\infty}H_{2n-1}\frac{\left(-1\right)^{n+1}}{n}=\frac{\pi^{2}}{16}-\frac{1}{4}\ln^{2}\left(2\right)$$

where the $\pi^2$ comes from taking the square of the logarithm of a complex value. Subtracting these from each other gets the desired result.