Finding $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$ I'm trying to find $\lim_{n\to\infty}(\frac{1}{\sqrt{n^2+1}} + \frac{1}{\sqrt{n^2+2}} + ... + \frac{1}{\sqrt{n^2+n}})$.


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*I tried to use the squeeze theorem, failed.

*I tried to use a sequence defined recursively: $a_{n+1} = {a_n} + \frac{1}{\sqrt{(n+1)^2 +n+1}}$. It is a monotone growing sequence, for every $n$, $a_n > 0$. I also defined $f(x) = \frac{1}{\sqrt{(x+1)^2 +x+1}}$. So $a_{n+1} = a_n + f(a_n)$. But I'm stuck.


How can I calculate it?

If $k\in[1,n]$ then the difference between $\frac{1}{\sqrt{n^2+k}}$ and $\frac{1}{n}$ is rather small:
$$ 0\leq \frac{1}{n}-\frac{1}{\sqrt{n^2+k}} = \frac{k}{n\sqrt{n^2+k}(n+\sqrt{n^2+k})}\leq \frac{1}{2n^2} $$
hence $\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}$ tends to $1$ as $n\to +\infty$, since $\sum_{k=1}^{n}\frac{1}{n}=1$ and $0\leq\sum_{k=1}^{n}\frac{1}{2n^2}=\frac{1}{2n}$.
