Find the sum $\sum_{n = 1}^{\infty}(-1)^{n + 1}\log(1 + (1/n))$

Question

I started as follows $$\begin{aligned}S &= \sum_{n = 1}^{\infty}(-1)^{n + 1}\log\left(1 + \frac{1}{n}\right)\\ &= \sum_{n = 1}^{\infty}(-1)^{n + 1}\sum_{k = 1}^{\infty}(-1)^{k + 1}\frac{1}{k n^{k}}\\ &= \sum_{k = 1}^{\infty}\frac{(-1)^{k + 1}}{k}\sum_{n = 1}^{\infty}(-1)^{n + 1}\frac{1}{n^{k}}\end{aligned}$$ Let $$f(k) = \sum_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n^{k}}$$ and this is related to $\zeta(k)$ for $k > 1$. Clearly $$f(k) = (1 - 2^{1 - k})\zeta(k)$$ and $f(1) = \log 2$. It follows that we have $$S = \log 2 - \sum_{k = 2}^{\infty}(-1)^{k}\cdot\frac{(1 - 2^{1 - k})\zeta(k)}{k}$$ After this I am not aware how to proceed further. Also I have doubt whether the sums in $k, n$ can be interchanged (because the series involved are conditionally convergent) as I have done above but for the time being I have assumed it to be so.

Please help me out here.

Answer 1

Your series is just the logarithm of the Wallis product, hence you have: $$ S = \sum_{n=1}^{+\infty}(-1)^{n+1}\log\left(1+\frac{1}{n}\right) = \log\left(\frac{\pi}{2}\right).$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} S&=\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\ln\pars{1 + {1 \over n}} =\sum_{n = 1}^{\infty}\pars{-1}^{n + 1}\int_{0}^{1}{\dd x \over x + n} =\int_{0}^{1}\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n + x + 1}\,\dd x \\[2mm]&=\int_{0}^{1} \sum_{n = 0}^{\infty}\pars{{1 \over 2n + x + 1} - {1 \over 2n + x + 2}}\,\dd x =\int_{0}^{1} \sum_{n = 0}^{\infty}{1 \over \pars{2n + x + 2}\pars{2n + x + 1}}\,\dd x \\[3mm]&={1 \over 4}\int_{0}^{1} \sum_{n = 0}^{\infty}{1 \over \pars{n + x/2 + 1}\pars{n + x/2 + 1/2}}\,\dd x =\half\int_{0}^{1}\bracks{% \Psi\pars{{x \over 2} + 1} - \Psi\pars{{x \over 2} + \half}}\,\dd x \end{align} where $\ds{\Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ is the Digamma Function. $\ds{\Gamma\pars{z}}$ is the Gamma Function.

\begin{align} \color{#66f}{\large S} &=\left.\ln\pars{\Gamma\pars{x/2 + 1} \over \Gamma\pars{x/2 + 1/2}} \right\vert_{\ x\ =\ 0}^{\ x\ =\ 1} =\ln\pars{{\Gamma\pars{3/2} \over \Gamma\pars{1}}\, {\Gamma\pars{1/2} \over \Gamma\pars{1}}} =\ln\pars{\half\,\Gamma^{\,2}\pars{\half}} \\[3mm]&=\color{#66f}{\large\ln\pars{\pi \over 2}} \end{align}

We used the identities: $$ \Gamma\pars{1} = 1\,,\qquad\Gamma\pars{z + 1} = z\,\Gamma\pars{z}\,,\qquad \Gamma\pars{\half} = \root{\pi} $$