Problem with counting limit. I am obliged to count the
$$\lim_{n\rightarrow\infty} a_n$$
where
$$a_n=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}.$$
I know that this type of task, I should use squeeze theorem. I should find two sequences one that is smaller and one that is bigger. So I find $\large b_n=\frac{1}{n^2+1}$, $ b_n \le a_n$, but I have trouble in finding the bigger sequence $c_n \ge a_n$. I will be glad for any help.
 A: We will use the following: $$\sum_{k=1}^n k=\frac{n(n+1)}{2}.$$
Since $n\geq 1$, we have
$$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\geq\frac{1}{n^2+n}+\frac{2}{n^2+n}+\frac{3}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n(n+1)}{2(n^2+n)}$$
and
$$a_{n}=\frac{1}{n^2+1}+\frac{2}{n^2+2}+\frac{3}{n^2+3}+\cdots+\frac{n}{n^2+n}\leq\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\cdots+\frac{n}{n^2}=\frac{n(n+1)}{2n^2}.$$
By the squeeze theorem, $\lim_{n\rightarrow\infty} a_n=\frac{1}{2}$.
A: Hint: use, for $1\leq k\leq n$ :
$$\frac{k}{n^2+n}\leq \frac{k}{n^2+k}\leq \frac{k}{n^2}.$$
The idea is to remove $k$ from the denominator.
A: Hint : $a_n= n^2(H_{\frac{n}{n+1}}+1$) where $H$ is a harmonic series, which in this case for large $n$ is $\sim -\frac{1}{n+1}+\frac{1}{2(n+1)^2}$.
Can you handle from here? 
A: Each term in the expression for $a_n$ is less than or equal to $\displaystyle \frac{n}{n^2+n} = \frac{1}{n+1}$, so I think the upper bound would be $$a_n = \frac{1}{n^2+1} + \frac{2}{n^2+2} + \cdots + \frac{n}{n^2+n} \le \frac{n}{n^2+n} + \frac{n}{n^2+n} + \cdots + \frac{n}{n^2+n} = \frac{n}{n+1} = c_n.$$
Of course, as $n \rightarrow \infty$, $c_n \rightarrow 1$.
