Find $\sum \limits_{n=1}^{\infty } (-1)^{n+1} \log (k+n)$ Could you find the followings sum
$$\sum \limits_{n=1}^{\infty } (-1)^{n+1} \log (k+n)$$
Thanks
 A: This series is divergent as proved by others but may be 'regularized'.
Let's use zeta regularization for that purpose and consider the function :
\begin{align}
\tag{1}f_k(s)&:=\sum\limits_{n=1}^{\infty } \frac{(-1)^n}{(k+n)^s}\\
&=\sum \limits_{m=0}^{\infty } \frac 1{(k+2+2m)^s}-\sum \limits_{m=0}^{\infty } \frac 1{(k+1+2m)^s}\\
&=2^{-s}\left[\sum \limits_{m=0}^{\infty } \frac 1{\left(\frac{k+2}2+m\right)^s}-\sum \limits_{m=0}^{\infty } \frac 1{\left(\frac{k+1}2+m\right)^s}\right]\\
\\
\tag{2}f_k(s)&=2^{-s}\left[\zeta\left(s,\frac{k+2}2\right)-\zeta\left(s,\frac{k+1}2\right)\right]\\
\end{align}
with $\zeta(s,n)$ the Hurwitz zeta function.
The equality $(2)$ may be shown valid for $\,\Re(s)>0\,$ using the mean value theorem but let's just suppose $\,\Re(s)>1$. By analytic continuation we may extend it on the whole complex plane (except at singular points of $\zeta\;$).
Using $\ \dfrac d{ds}(k+n)^{-s}=-\dfrac{\log(k+n)}{(k+n)^s}\;$ we may then define the regularized sum as :
\begin{align}
\tag{3}S_k&:=\lim_{s\to 0}\;\frac d{ds}\;f_k(s)=\sum \limits_{n=1}^{\infty } (-1)^{n+1} \log (k+n)\\
\tag{4}S_k&=\left.\frac d{ds}\right|_{s=0}2^{-s}\left[\zeta\left(s,\frac{k+2}2\right)-\zeta\left(s,\frac{k+1}2\right)\right]\\
\end{align}
This may be further simplified using this and this formula from Wolfram functions :
$$\left.\frac d{ds}\right|_{s=0}2^{-s}=-\log(2),$$
$$\left.\frac d{ds}\right|_{s=0}\zeta\left(s,a\right)=\log\,\Gamma(a)+\zeta'(0)=\log\,\Gamma(a)-\frac{\log(2\pi)}2,$$
$$\lim_{s\to 0}\;\zeta\left(s,a\right)=\frac 12-a,$$
Putting all this together allows to evaluate $(4)$ :
\begin{align}
S_k&=-\log(2)\left[-\frac {k+1}2+\frac k2\right]+\left[\log\,\Gamma\left(\frac{k+2}2\right)-\log\,\Gamma\left(\frac{k+1}2\right)\right]\\
S_k&=\frac{\log(2)}2+\log\,\frac{\Gamma\left(\frac{k+2}2\right)}{\Gamma\left(\frac{k+1}2\right)}\\
\end{align}
And the general formula for any complex $k$ not a negative integer will be :
$$\tag{5}\boxed{\displaystyle S_k=\log\frac{\sqrt{2}\;\Gamma\left(\frac {k+2}2\right)}{\Gamma\left(\frac{k+1}2\right)}}$$
A: Assuming $k$ is a fixed number, this series diverges, since the terms don't tend to $0$.
A: \begin{align}
\sum_{n = 1}^{\infty}\left(-1\right)^{n + 1}\ln\left(k + n\right)
&=
\sum_{n = 1}^{\infty}\left\lbrack%
\ln\left(k + 2n - 1\right) - \ln\left(k + 2n\right)
\right\rbrack
=
\sum_{n = 1}^{\infty}\ln\left(k + 2n - 1 \over k + 2n\right)
\\[3mm]&=
\sum_{n = 1}^{\infty}\ln\left(1 - {1 \over k + 2n}\right)
\quad
\left\vert%
\begin{array}{l}{\rm
The\ general\ term\ behaves}\ \sim -\,{1 \over 2}\ {1 \over n}
\\
{\rm when}\ n \gg 1\ \Longrightarrow\ {\rm diverges!!!.} 
\end{array}\right.
\end{align}
