Calculate: $\lim\limits_{n\to\infty} \sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}$ Calculate: $$\lim\limits_{n\to\infty} \sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}$$
I thought a Riemann sum could lead to something, but couldn't find a suitable partition.
Hint, please?
 A: One may recognize a Riemann sum, by writing
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac1n \cdot\sum_{k=0}^{n} \frac{2+\frac{k}n }{1+(2+\frac{k}n)^2},
$$ then letting $n \to \infty$, to obtain
$$
\frac1n \cdot\sum_{k=0}^{n} \frac{2+\frac{k}n}{1+(2+\frac{k}n)^2} \to \int_0^1 \frac{2+x }{1+(2+x)^2}\:dx.\tag1
$$
Add-on. Since $f:[0, 1] \rightarrow [0, 1]$ with $f(x)=\frac{2+x}{1+(2+x)^2}$ satisfies $f \in \mathcal{C}^1([0,1])$, then one is allowed to apply the standard result
$$
\frac1n\sum_{k=0}^{n} f\left(\frac{k}{n}\right) =\int_0^1 f(x)\,dx + \frac{f(0) + f(1)}{2n}+o\left(\frac1n \right) \tag2
$$ giving, as $n \to \infty$,

$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac{\ln 2}2+\frac{7}{20\: n}+o\left(\frac1n \right). \tag3
$$

One may in fact express the given sum in terms of the digamma function, using
$$
\sum_{k=0}^{n} \frac{1}{k+b}=\psi\left(n+b+1\right)-\psi\left(b\right), \qquad \text{Re}\:b>0,
$$ and writing
$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2} = \text{Re}\:\sum_{k=0}^{n} \frac{1}{k+(2+i)n}
$$ then recalling the asymptotics of the digamma function, as $n \to \infty$, one obtains

$$
\sum_{k=0}^{n} \frac{2n+k}{n^2+(2n+k)^2}=\frac{\ln 2}2+\frac7{20\:n}+\frac1{300\:n^2}+\frac7{60\:000 \:n^4}+o\left(\frac1{n^5} \right).\tag4
$$

