How can we one show that $\sum_{i=1}^{\infty}\sum_{j=1}^{2k}(-1)^{j-1}{2\over i+j}=H_k?$

Question

Given the double sums $(1)$

$$\sum_{i=1}^{\infty}\sum_{j=1}^{2k}(-1)^{j-1}{2\over i+j}=\color{blue}{H_k}\tag1$$ Where $H_k$ is the n-th harmonic number

How can one prove $(1)$?

Rewrite $(1)$ as

$$\sum_{i=1}^{\infty}\left({2\over i+1}-{2\over i+2}+{2\over i+3}-\cdots+{2\over i+k}\right)\tag2$$

Rewrite $(2)$ as

$$\sum_{i=1}^{\infty}\left({2\over (i+1)(i+2)}+{2\over (i+3)(i+4)}+{2\over (i+5)(i+6)}+\cdots+{2\over (i+k)(i+k+1)}\right)\tag3$$

Help required, not sure what is the next step. Thank you.

Answer 1

There is some issue: if $k$ is odd, the main term of $(2)$ behaves like $\frac{C}{i}$, leading to a divergent sum.

On the other hand, if $k=2h$ is even then $(2)$ is an absolutely convergent telescopic sum:

$$ \sum_{i\geq 1}\left(\frac{2}{i+1}-\frac{2}{i+2}+\frac{2}{i+3}-\frac{2}{i+4}+\ldots+\frac{2}{i+2h-1}-\frac{2}{i+2h}\right) $$ that clearly equals $$ \frac{2}{1+1}+\frac{2}{1+3}+\ldots+\frac{2}{2h} = H_h. $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{i = 1}^{\infty}\sum_{j = 1}^{2k}\pars{-1}^{\,j - 1}{2 \over i + j} & = 2\sum_{i = 1}^{\infty} \sum_{j = 1}^{2k}\pars{-1}^{\,j - 1}\int_{0}^{1}x^{i + j - 1}\,\dd x = 2\sum_{i = 1}^{\infty}x^{i} \int_{0}^{1}\sum_{j = 1}^{2k}\pars{-x}^{\,j - 1}\,\dd x \\[5mm] & = 2\sum_{i = 1}^{\infty}x^{i} \int_{0}^{1}{\pars{-x}^{2k} - 1 \over -x - 1}\,\dd x = 2\int_{0}^{1}{1 - x^{2k} \over 1 + x}\sum_{i = 1}^{\infty}x^{i}\,\dd x \\[5mm] & = 2\int_{0}^{1}{x - x^{2k + 1} \over 1 - x^{2}}\,\dd x \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, \int_{0}^{1}{x^{k} - 1 \over x - 1}\,\dd x = \int_{0}^{1} \sum_{n = 1}^{k}x^{n - 1}\,\dd x \\[5mm] & = \sum_{n = 1}^{k}{1 \over n} = \bbx{\ds{H_{k}}} \end{align}