2
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $$ \frac{2}{1+1}+\frac{2}{1+3}+\ldots+\frac{2}{2h} = H_h. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, there was a typo I did. $\endgroup$ – gymbvghjkgkjkhgfkl Mar 28 '17 at 4:36
1
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.