Find the sum $\sum_{n = 1}^{\infty}(-1)^{n + 1}\log(1 + (1/n))$ 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.
 A: Your series is just the logarithm of the Wallis product, hence you have:
$$ S = \sum_{n=1}^{+\infty}(-1)^{n+1}\log\left(1+\frac{1}{n}\right) = \log\left(\frac{\pi}{2}\right).$$
A: $\newcommand{\+}{^{\dagger}}
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\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:
$$
\Gamma\pars{1} = 1\,,\qquad\Gamma\pars{z + 1} = z\,\Gamma\pars{z}\,,\qquad
\Gamma\pars{\half} = \root{\pi}
$$
