Reading a [paper](https://www.sciencedirect.com/science/article/pii/0001870874900218) about eta invariants I came across a zeta-like function.


I'm looking for the analytic continuation of $$\sum\_{k=1}^\infty k(k+a)^{-s}$$ at $s=0$, where $a$ is positive.


In the paper he just says "The [...] term causes no problem and at $s=0$ has the value $\left[\frac{(4a^2-1)}{12}\right]$." Unfortunately, I really don't see the that.


My first approach so far; I tried a Taylor series at $s=3$:
\begin{align}
\sum\_{k=1}^\infty k(k+a)^{-s} &= \sum\_{k=1}^\infty k \sum\_{l=0}^\infty (-1)^l \begin{pmatrix}2+l \\ l\end{pmatrix} (k+a)^{-3-l} (s-3)^{l}
\end{align}
and inserted $s=0$
$$\sum\_{k=1}^\infty \frac{k}{(k+a)^3} \sum\_{l=0}^\infty \begin{pmatrix}2+l \\ l\end{pmatrix} \left( \frac{3}{k+a} \right)^l$$
which is for $a\geq 3$
$$ \sum\_{k=1}^\infty \frac{k}{(k+a-3)^3}~. $$
That seems to converge if I didn't miscalculate. But what are the further steps in order to get the result above? Or is there a more skilful approach?


Thanks in advance!



Analytic Continuation of Zeta-like function
Let $a>0$. We can write
$$f(s,a):=\sum\_{k=1}^\infty k(k+a)^{-s}=\sum\_{k=0}^\infty (k+a)^{-s+1}-a\sum\_{k=0}^\infty (k+a)^{-s}=\zeta(s-1,a)-a\zeta(s,a).$$
Hence, $f$ has a meromorphic continuation to $\mathbb{C}$ with simple poles at $s=1$ and $s=2$. 


Now, if $s=-n$ is a non-positive integer, it is known that
$$\zeta(-n,a)=-\frac{B\_{n+1}(a)}{n+1}$$
where $B\_n(X)$ is the $n$-th Bernoulli polynomial. Hence,
$$f(0,a)=\zeta(-1,a)-a\zeta(0,a)=-\frac{B\_{2}(a)}{2}+aB\_1(a).$$ 
Since $B\_1(X)=X-1/2$ and $B\_2(X)=X^2-X+1/6$, we obtain that
$$f(0,a)=-\frac{6a^2-6a+1}{12}+a\frac{2a-1}{2}=\frac{6a^2-1}{12},$$
(unless I made some stupid mistake which I don't have the time to fix now).


