On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$

Question

I managed to prove through complex analysis that

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

However, I'm having a difficult time proving this result with real analysis methods. Partial summation of the series gets nasty pretty quickly since it involves harmonic numbers. Another interesting fact to note about this series is that this part

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

diverges. This came as a complete surprise to me since I expected this to telescope. What remains now to prove the series I want is by using generating function.

\begin{align*} \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )x^n &=\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{2n-1} - \sum_{n=1}^{\infty} \frac{x^n}{2n+1} \\ &= -\log \left ( 1-x \right ) - \sqrt{x} \;\mathrm{arctanh} \sqrt{x} - \frac{{\mathrm {arctanh} }\sqrt{x} - \sqrt{x}}{\sqrt{x}} \end{align*}

I cannot , however, evaluate the limit as $x \rightarrow 1^-$ of the last expression. Can someone finish this up?

Of course alternatives are welcome.

Answer 1

A very simple way: $$ \begin{eqnarray*}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}\right)&=&\sum_{n\geq 1}\int_{0}^{1}\left( 2x^{2n-1}-x^{2n-2}-x^{2n}\right)\,dx\\&=&\int_{0}^{1}\sum_{n\geq 1}x^{2n-2}(2x-1-x^2)\,dx\\&=&\int_{0}^{1}\frac{2x-1-x^2}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x-1}{x+1}\,dx=\left[x-2\log(1+x)\right]_{0}^{1}=\color{red}{1-2\log 2}.\end{eqnarray*} $$

Answer 2

$$\sum_{n\ge 1}\left(\frac1n - \frac1{2n-1} - \frac1{2n+1}\right) = \sum_{n\ge 1}\left(\frac1{2n} - \frac1{2n-1}\right) + \sum_{n\ge 1}\left(\frac1{2n} - \frac1{2n+1}\right) = \sum_{n \ge 1} \frac{(-1)^n}{n} + \sum_{n\ge 2}\frac{(-1)^n}{n} = 2\sum_{n\ge 1} \frac{(-1)^n}{n} + 1 = -2\log(1-(-1)) + 1 = 1 - 2\log2$$

Answer 3

Notice $$\sum_{n=1}^p \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right) = \sum_{n=1}^p \left[\frac2n - \left( \frac{1}{2n-1} + \frac{2}{2n} + \frac{1}{2n+1} \right)\right]\\ = 2H_p - \left(1 + \sum_{n=2}^{2p} \frac{2}{n} + \frac{1}{2p+1}\right) = 2(H_p - H_{2p}) + 1 - \frac{1}{2p+1}$$

where $H_p = \sum_{n=1}^p \frac{1}{n}$ is the $p^{th}$ harmonic number. Since $\lim\limits_{p\to\infty} (H_p - \log p) = \gamma$, we have $$\begin{align}\lim_{p\to\infty} (H_p - H_{2p}) &= \lim_{p\to\infty} \left((H_p - \log p) - (H_{2p} - \log(2p)) - \log 2\right)\\ &= \gamma - \gamma - \log 2 = -\log 2\end{align}$$

As a result,

$$\sum_{n=1}^\infty \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right) = 2(-\log 2)+1 - 0 = 1 - 2\log 2$$

Answer 4

In the same spirit as achille hui, $$S_p=\sum_{n=1}^p \left(\frac1n - \frac{1}{2n-1} - \frac{1}{2n+1}\right)$$ $$(2p+1)S_p=2 \gamma p+2 p-2 p \psi ^{(0)}\left(p+\frac{1}{2}\right)+2 p \psi ^{(0)}(p+1)+2 p \psi ^{(0)}\left(\frac{1}{2}\right)-\psi ^{(0)}\left(p+\frac{1}{2}\right)+\psi ^{(0)}(p+1)+\gamma +\psi ^{(0)}\left(\frac{1}{2}\right)$$ where appears the digamma function.

Using generalized harmonic numbers, this reduces to $$S_p=H_p-H_{p-\frac{1}{2}}+\frac{2 p}{2 p+1}-2\log (2)$$ Now, using the asymptotics $$H_q=\gamma +\log \left({q}\right)+\frac{1}{2 q}-\frac{1}{12 q^2}+O\left(\frac{1}{q^4}\right)$$ and continuing with Taylor expansion $$S_p=1-2\log (2)+\frac{1}{8 p^2}-\frac{1}{8 p^3}+O\left(\frac{1}{p^4}\right)$$ which allows quite accurate results even for small values of $p$.

For example $S_5=-\frac{5297}{13860}\approx -0.382179$ while the above expansion would give $\frac{251}{250}-2\log (2)\approx -0.382294$.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mbox{Note that}\quad\left.\sum_{n = 1}^{N}{1 \over an + b}\,\right\vert_{\ a\ \not=\ 0} & = {1 \over a}\sum_{n = 1}^{N}{1 \over n + b/a} = {1 \over a}\sum_{n = 1}^{\infty}\pars{{1 \over n + b/a} - {1 \over n + N + b/a}} \phantom{\mbox{Note that}} \\[5mm] & = {H_{N + b/a} - H_{b/a} \over a}\qquad \pars{~H_{z}:\ Harmonic\ Number~} \end{align}

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\begin{align} &\bbox[10px,#ffd]{\ds{% \sum_{n = 1}^{\infty}\pars{{1 \over n} - {1 \over 2n - 1} - {1 \over 2n + 1}}}} = \lim_{N \to \infty}\pars{H_{N} - {H_{N - 1/2} - H_{-1/2} \over 2} - {H_{N + 1/2} - H_{1/2} \over 2}} \\[5mm] = &\ {1 \over 2}\,\lim_{N \to \infty}\bracks{\ln\pars{N^{2}} - \ln\pars{N - {1 \over 2}} - \ln\pars{N + {1 \over 2}}} + {1 \over 2}\int_{0}^{1}\pars{{1 - t^{-1/2} \over 1 - t} + {1 - t^{1/2} \over 1 - t}}\,\dd t \\[5mm] = &\ -\,{1 \over 2}\,\lim_{N \to \infty}\ln\pars{\bracks{1 - {1 \over 2N}}\bracks{1 + {1 \over 2N}}} + {1 \over 2}\int_{0}^{1}{2 - 1/t - t \over 1 - t^{2}}\,2t\,\dd t = -\int_{0}^{1}{t^{2} -2t + 1 \over 1 - t^{2}}\,\dd t \\[5mm] = &\ \int_{0}^{1}{t - 1 \over 1 + t}\,\dd t = \int_{0}^{1}\dd t - 2\int_{0}^{1}{\dd t \over 1 + t} = \bbx{1 - 2\ln\pars{2}} \approx -0.3863 \end{align}

Note that $\ds{H_{z} = \int_{0}^{1}{1 - t^{z} \over 1 - t}\,\dd t\,, \quad\Re\pars{z} > - 1}$ and

$\ds{H_{z} \,\,\,\stackrel{\mrm{as}\ \verts{z}\ \to\ \infty}{=}\,\,\,\ln\pars{z} + \gamma + {1 \over 2z} + \,\mrm{O}\pars{1 \over z^{2}}}$ where $\ds{\verts{\arg\pars{z}} < \pi}$. $\ds{\gamma}$ is the Euler-Mascheroni Constant.