Calculate $ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+...+\frac{1}{\sqrt{n^{2}+n}}\right) $

Question

I have to calculate the limit :

$ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+...+\frac{1}{\sqrt{n^{2}+n}}\right) $

Now here's what I've tried:

$ \lim_{n\to\infty}\left(\frac{1}{\sqrt{n^{2}+1}}+\frac{1}{\sqrt{n^{2}+2}}+...+\frac{1}{\sqrt{n^{2}+n}}\right)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n^{2}+k}}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^{2}}}}$

Now to me it looks like a Riemman sum of the function $ f\left(x\right)=\frac{1}{\sqrt{1+x^{2}}} $, the partition $ P_{n}=\left\{ 0,\frac{1}{n},\frac{2}{n},...,\frac{n}{n}\right\} $ and for the interval $ [\frac{j-1}{n},\frac{j}{n}] $ we choose that point $ \frac{\sqrt{j}}{n} $ so the sequence of points would be $ \theta_{n}=\left\{ \frac{\sqrt{1}}{n},\frac{\sqrt{2}}{n},...,\frac{\sqrt{n}}{n}\right\} $.

Thus

$ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n}\frac{1}{\sqrt{1+\frac{k}{n^{2}}}}=\intop_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx=\ln\left(|\sqrt{x^{2}+1}+x|\right)|_{0}^{1}=\ln\left(\sqrt{2}+1\right) $

But I doubt myself because the calculation of the integral $ \intop_{0}^{1}\frac{1}{\sqrt{1+x^{2}}}dx $ is not very easy and Im not sure if this is the right way.

So I'd appreciate if someone can tell if this solution holds.

If my solution is not correct I'd happy to see a correct solution.

Answer 1

The solution is incorrect because in general $$\frac{\sqrt{j}}{n} \notin [\frac{j-1}{n},\frac{j}{n}] $$

To find this limit notice that for $1\le k\le n$ $$ \frac{1}{\sqrt{n^2+n}} \le \frac{1}{\sqrt{n^2+k}} \le \frac{1}{n} $$ so $$ \frac{1}{\sqrt{1+\frac{1}{n}}}=\frac{n}{\sqrt{n^2+n}} \le \sum_{k=1}^n\frac{1}{\sqrt{n^2+k}} \le n \cdot \frac{1}{n} =1$$ And use the squeeze theorem.