Computing: $\lim\limits_{n\to\infty}\sum\limits_{k=1}^{\infty} \frac{n}{nk^2+k+1}$ I would like to find an equivalence at infinity  to the following sequence.
$$C_n= \sum_{k=1}^{\infty} \frac{n}{nk^2+k+1}$$
Here are the given questions

1-Find the $\lim_{n\to\infty}C_n$
2- If the limit blows up give an equivalence at infinity.

For the first question, I set $$C_n(k)=  \frac{n}{nk^2+k+1}$$
then $$\lim_{n\to\infty}C_n(k) =\frac{1}{k^2} $$ and I proved that the sequence $(C_n(k))_n$ is monotone
it therefore springs from monotone convergence
that,
$$\lim_{n\to\infty}C_n=\lim_{n\to\infty}\sum_{k=1}^{\infty} \frac{n}{nk^2+k+1}= \sum_{k=1}^{\infty} \lim_{n\to\infty}\frac{n}{nk^2+k+1} = \sum_{k=1}^{\infty} \frac{1}{k^2} =\frac{\pi^2}{6}$$
Is this reasoning  correct if yes then I am interested on knowing some more elegant way (if there is some) of solving this. if not give me answer. I am don't how to solve the question in the case the limit is infinity.
 A: $$\sum_{k\geq 0}\frac{1}{(k+a)(k+b)}=\frac{\psi(a)-\psi(b)}{a-b}\qquad\left(\scriptstyle{\psi(x)=\frac{d}{dx}\log\Gamma(x)}\right)$$
hence by factoring $k^2+\frac{k}{n}+\frac{1}{n}$
$$\sum_{k\geq 1}\frac{n}{n k^2+k+1}= \frac{\pi^2}{6}-\frac{\zeta(3)+\zeta(4)}{n}+O\left(\frac{1}{n^2}\right).$$
Needless to say, the first terms of the asymptotic expansion can be found by creative telescoping™ too.
A: HINT:
The first term of the series is $n$.  For the remaining part, note that we have
$$\begin{align}
\left|\sum_{k=1}^\infty \left(\frac1{k^2}-\frac{1}{k^2+\frac{k+1}{n}}\right)\right|&=\frac1n\sum_{k=1}^\infty \frac{k+1}{k^2\left(k^2+\frac{k+1}{n}\right)}\\\\&\le \frac2n \sum_{k=1}^\infty \frac{1}{k^3}
\end{align}$$
A: From integration point of view, $C_n=\int_{\Bbb N} f(n,k) d m(k)$ with $f(n,k)= \frac{n}{nk^2+k+1}$ where $dm$ is counting measure: m(A)=# A= card (A). We clearly have that
$$ \lim_{n\to \infty} f(n,k)= g(k)=\frac{1}{k^2}\quad \text{and}\quad f(n,k)\leq g(k)=\frac{1}{k^2}\in L^1(\Bbb N, dm)$$
since
$$\int_{\Bbb N} g(k) d m(k)=\sum_{k\in\Bbb N} \frac{1}{k^2}=\frac{\pi^2}{6}.$$
By convergence dominated theorem we have
$$ \lim_{n\to \infty} C_n= \int_{\Bbb N} \lim_{n\to \infty} f(n,k) d m(k)=\int_{\Bbb N}g(k)d m(k)=\sum_{k\in \Bbb N} \frac{1}{k^2}=\frac{\pi^2}{6}$$
