A series involving alternating Harmonic numbers

Calculus (Integrals, Series)
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mathofusva
Posts: 33
Joined: Tue May 10, 2016 3:56 pm

A series involving alternating Harmonic numbers

#1

Post by mathofusva »

For $k \in \mathbb{N}$, let
$$S(k) = \sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+k}\,H_n,$$
where $H_n$ is the $n$-th harmonic number. It is known that
$$S(0) = \frac{\pi^2}{12} - \frac{1}{2}\,\ln^22,\,\,\,\, S(1) = \frac{1}{2}\,\ln^22.$$
Can you find a closed form for $S(k)$ in general?
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Re: A series involving alternating Harmonic numbers

#2

Post by Riemann »

We are proving that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \mathcal{H}_n}{n} = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$. Indeed, note that:

\begin{align*}
\sum_{n=1}^N\frac{(-1)^{n-1}}{n}H_n
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\sum_{n=2}^N\frac{(-1)^{n-1}}{n}H_{n-1}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{n}\left(\frac1k+\frac1{n-k}\right)\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{n=2}^N\sum_{k=1}^{n-1}\frac{(-1)^{n-1}}{k(n-k)}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=k+1}^N\frac{(-1)^{n-1}}{k(n-k)}\\
&=\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}+\frac12\sum_{k=1}^{N-1}\sum_{n=1}^{N-k}\frac{(-1)^{n+k-1}}{kn}\\
&=\color{#00A000}{\sum_{n=1}^N\frac{(-1)^{n-1}}{n^2}}
-\color{#0000FF}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=1}^{N-1}\frac{(-1)^{n-1}}{n}}\\
&\quad \quad \quad +\color{#C00000}{\frac12\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}}\tag{1}
\end{align*}

where, using the Alternating Series Test, we have

\begin{align*}
&\color{#C00000}{\frac12\left|\sum_{k=1}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}\right|}\\
&\le\frac12\left|\sum_{k=1}^{N/2}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}\right|
+\frac12\left|\sum_{k=N/2}^{N-1}\frac{(-1)^{k-1}}{k}\sum_{n=N-k+1}^{N-1}\frac{(-1)^{n-1}}{n}\right|\\
&\le\frac12\cdot1\cdot\frac2N+\frac12\cdot\frac2N\cdot1\\
&=\frac2N\tag{2}
\end{align*}

The result follows.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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