<s>[INST] $\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, is my solution wrong? I needed to calculate: $$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$$
First of all I saw that it won't be possible to do that in any traditional way and actually calculate the limit, because of the form of expression. I mean - it's a sum with sqares on $n$ so I can't use Stolz lemma that easy. But, I thoght, that the solution is probably $0$, because probably every element of the sum is $0$ when $n \implies \infty$ and the limitation of sum in $\infty$ = sum of limitations in $\infty$. So I just went with that and decided to prove that using induction.
My base is:
$$\lim_{n \to \infty} \frac{1}{\sqrt{n^2 + 1}} = 0$$
My assumption:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +...+ \frac{1}{\sqrt{n^2 +2n}}\right) = 0$$
My induction:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0 + \lim_{n \to \infty} \frac{1}{\sqrt{n^2 +2n + 1}}) = 0$$
So the limit is:
$$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right) = 0$$
But then my grader at university said that at first sight it looks totaly wrong but he actualy needs to think about it a bit. So here is my question - is that wrong? How is that wrong?
Ok, thank you for your answers. I thought I can solve that exercise that way because I asked not long ago very similar question on that forum: Find the limit of of such a sequence defined by recurrence
Because of your answers I think that the problem is actually that in this case I am dealing with a SUM of elements, am I right (or it the answer that I got in other case wrong?)?
 [/INST] Clearly,
$$
\frac{1}{n+1}\le \frac{1}{\sqrt{n^2 + k}} \le \frac{1}{n}
$$
for $k=1,2,\ldots,2n+1$. Hence
$$
\frac{2n+1}{n+1}<\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}<\frac{2n+1}{n}
$$
and
$$
\lim_{n\to\infty}\frac{2n+1}{n+1}=\lim_{n\to\infty}\frac{2n+1}{n}=2.
$$
Therefore
$$
\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\to 2.
$$
</s>