Convergence of $S(x,y)=\sum_{n=1}^{\infty}\left(\frac{x+2n+1}{(x+2n+1)^2+y^2}+\frac{x+2n}{(x+2n)^2+y^2}-\frac{x+n}{(x+n)^2+y^2}\right)$ I know that the following sum is convergent and equal to $\ln2$ for all $x\not\in\mathbb{Z}$
$$\sum_{n=0}^{\infty}\left(\frac{1}{x+2n+1}+\frac{1}{x+2n}-\frac{1}{x+n}\right)-\ln2=0$$
Can I somehow quickly prove that the following sum is also convergent for $x\not\in\mathbb{Z}$ and $y\in\mathbb{R}$ ?
$$S(x,y)=\sum_{n=0}^{\infty}\left(\frac{x+2n+1}{(x+2n+1)^2+y^2}+\frac{x+2n}{(x+2n)^2+y^2}-\frac{x+n}{(x+n)^2+y^2}\right)$$
 A: For me, it is easier to consider the difference of the sums and use $$\frac{1}{\Omega}-\frac{\Omega}{\Omega^2+y^2}=\frac{y^2}{\Omega(\Omega^2+y^2)},$$ with the convergence seen immediately then.
In fact, since $\sum_{n=0}^{m-1}(a_{2n+1}+a_{2n}-a_n)=\sum_{n=m}^{2m-1}a_n$, partial sums of both series are negligibly different from Riemann sums. Namely, for $a_n=1/(x+n)$, we have (say, throwing in/out a finite number of terms) $$\sum_{n=m}^{2m-1}\frac{1}{x+n}=\frac{1}{m}\sum_{n=0}^{m-1}\frac{1}{1+\frac{n+x}{m}}\underset{m\to\infty}{\longrightarrow}\int_0^1\frac{dt}{1+t}=\ln 2$$ as you know already, and a similar argument shows that $\color{blue}{S(x,y)=\ln 2}$ for all allowed $x,y$.
A: According to Wolfy,
each term is
$
\dfrac{\left(8 n^4 x - 4 n^4 + 20 n^3 x^2 - 4 n^3 x + 12 n^3 y^2 - 2 n^3 + 18 n^2 x^3 + 3 n^2 x^2 + 18 n^2 x y^2 - 3 n^2 x + 9 n^2 y^2 + 7 n x^4 + 4 n x^3 + 10 n x^2 y^2 - n x^2 + 8 n x y^2 + 3 n y^4 + n y^2 + x^5 + x^4 + 2 x^3 y^2 + 2 x^2 y^2 + x y^4 + y^4\right)}
{(n^2 + 2 n x + x^2 + y^2) (4 n^2 + 4 n x + x^2 + y^2) (4 n^2 + 4 n x + 4 n + x^2 + 2 x + y^2 + 1)}
=\dfrac{\left(n^4(8 x - 4) + n^3(20  x^2 - 4  x + 12  y^2 - 2)\\
 + n^2(18  x^3 + 3  x^2 + 18  x y^2 - 3  x + 9  y^2 )\\
+ n(7 x^4 + 4  x^3 + 10 x^2 y^2 -  x^2 + 8  x y^2 + 3  y^4)\\ +  y^2 + x^5 + x^4 + 2 x^3 y^2 + 2 x^2 y^2 + x y^4 + y^4\right)}
{\left((n^2 + 2 n x + x^2 + y^2) \\(4 n^2 + 4 n x + x^2 + y^2) \\(4 n^2 + 4 n x + 4 n + x^2 + 2 x + y^2 + 1)\right)}
$.
Since the highest power of $n$
in the numerator is
$n^4$
and the denominator has
$n^6$,
the terms are like
$\dfrac1{n^2}$
and the sum of these converge.
