# Calculate the limit $\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$

I have to calculate the following limit: $$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$$

I believe the limit equals $1$, and I think I can prove it with the squeeze theorem, but I don't really know how.

Any help is appreciated, I'd like to receive some hints if possible.

Thanks!

For every $n>0$,
$$\frac{n}{\sqrt{n^2+n}}\le\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\le\frac{n}{\sqrt{n^2+1}}$$

Can you continue with Squeeze theorem?

• Got it. Great, thanks. I'll post the answer in full, later, as well for future users.
– Alan
Commented Jan 27, 2016 at 6:43
• Can you maybe explain why the expression in the middle is smaller and bigger than the left and right expressions?
– Alan
Commented Jan 28, 2016 at 4:42
• Because n->inf, we must have a N>0 s.t. n>N. Observe the expression in the middle, it is obvious that the maximum term is $\frac{1}{\sqrt{n^2+1}}$ with the condition n>N , if we replace all the term with the maximum term, it is obviously greater than the middle expression.The left expression follow the same idea. Commented Jan 28, 2016 at 5:17

Note that we can write the term of interest as

$$\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}=\frac1n\sum_{k=1}^n\frac{1}{\sqrt{1+k/n^2}}$$

To evaluate the limit we can expand the summand using the binomial theorem as

$$\frac{1}{\sqrt{1+k/n^2}}=1-\frac{k}{2n^2} +O(k^2/n^4)$$

Therefore, we have

\begin{align} \lim_{n\to \infty}\sum_{k=1}^n\frac{1}{\sqrt{n^2+k}}&=\lim_{n\to \infty}\frac1n \sum_{k=1}^n \left(1+O(k/n^2)\right)\\\\ &=\lim_{n\to \infty}\frac1n\left(n+O(1)\right)\\\\ &=1 \end{align}

• I don't understand the method 2, could you elaborate a bit? Commented Jan 27, 2016 at 5:21
• The problem with this approach is that, in Riemann sums, the function one considers the integral of, should not depend on $n$. A cute exercise would be to find examples where $$\lim_{n\to\infty}\frac1n\sum_{k=1}^nf_n\left(\frac{k}n\right)\ne\int_0^1\lim_{n\to\infty}f_n(x)\,dx.$$
– Did
Commented Jan 27, 2016 at 6:55
• @Did While the approach will generally not be tractable, the development is valid in this example, is it not? I am happy to delete if you think it might mislead. Commented Jan 27, 2016 at 15:13

This is the final answer I got, thanks to all the help:

For every $n>0$,$\frac{n}{\sqrt{n^2+n}}\le \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right) \le \frac{1}{\sqrt{n^2+n}}$

Using the squeeze theorem, we calculate the middle expression's limit.

$\lim_\limits{n \to \infty} \frac{n}{\sqrt{n^2+n}}=\lim_\limits{n \to \infty} \frac{\sqrt{n^2}}{\sqrt{n^2+n}}= \lim_\limits{n \to \infty}\sqrt{\frac{ {n^2}}{{n^2+n}}}=\lim_\limits{n \to \infty}\sqrt{\frac{ \frac{n^2}{n^2}}{{\frac{n^2}{n^2}+\frac{n}{n^2}}}}=\lim_\limits{n \to \infty} \sqrt{\frac{1}{1+\frac{1}{n}}}=1$ from limits arithmetic.

Likewise, we can calculate the right hand side and reach to the conclusion that the original sequence approaches $1$.