$\lim_{n \to \infty} \left(\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\right)$, 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?)?
 A: 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
$$
\frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} +\cdots+ \frac{1}{\sqrt{n^2 +2n + 1}}\to 2.
$$
A: The others already told you why your solution is wrong. I am going to show you how to actually compute this limit (and please bear in mind that this is a standard trick for such limits).
Obviously, $\displaystyle n^2+1\le n^2+k \le n^2+2n+1$ for any $n \in \mathbb{N}$, $k= \overline{1,2n+1}$. This is equivalent to saying that $\displaystyle \frac{1}{(n+1)^2}\le \frac{1}{n^2+k}\le \frac{1}{n^2+1}$. After you take the square root and sum up all the inequlities you get that $\displaystyle \frac{2n+1}{n+1}\le \sum_{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}\le \frac{2n+1}{\sqrt{n^2+1}}, \forall n\in \mathbb{N}$ and now by the squeeze theorem you may conclude that $\displaystyle \lim_{n\to \infty}\sum_{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}=2.$
A: By an induction like this you can correctly prove that for any constant $k$,
$$
   \lim_{n \to \infty} \left(\frac1{\sqrt{n^2+1}} + \frac1{\sqrt{n^2+2}} + \dots + \frac1{\sqrt{n^2+k}} \right) = 0.
$$
But there is no value of $k$ for which this sum of a fixed number of terms will turn into the sum in the question, which has a variable number of terms.
Going in the other direction, you can "bite off" any number of terms, and show that
$$
   \lim_{n \to \infty} \left(\frac1{\sqrt{n^2+1}} + \frac1{\sqrt{n^2+2}} + \dots + \frac1{\sqrt{n^2+2n+1}} \right) =    \lim_{n \to \infty} \left(\frac1{\sqrt{n^2+k}} + \frac1{\sqrt{n^2+k+1}} + \dots + \frac1{\sqrt{n^2+2n+1}} \right).
$$
But there is no value of $k$ which will reduce this to the limit that you've called the base case, because the number of terms will always keep growing with $n$.

In fact, since there are $2n+1$ terms in the sum, and each one of them is at least $\frac1{\sqrt{n^2+2n+1}} = \frac1{n+1}$, that the sum is equal to at least $\frac{2n+1}{n+1} = 2 - \frac1{n+1}$, and therefore the limit is at least $2$, not $0$. Think about what kind of upper bound we can put on the sum in the same way...
A: Using generalized harmonic numbers
$$a_n=\sum_{k=1}^{2n+1}\frac{1}{\sqrt{n^2+k}}=H_{n^2+2 n+1}^{\left(\frac{1}{2}\right)}-H_{n^2}^{\left(\frac{1}{2}\right)}$$
Using the asymptotics
$$H_p^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac{1}{2 \sqrt{p}}$$ and continuing with Taylor series
$$a_n=2-\frac{1}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
