Show that a special sequence converge How can I prove that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n + i}}$ converge. I know that if a sequence $(b_n)$ converge to $b$, then $(\sqrt{b_n})$ converge to $\sqrt{b}$ and $\frac{n}{\sqrt{2n}} \leq a_n \leq \frac{n}{\sqrt{n + 1}}$, but I am not sure how to finish this. Can you help?
Observe that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n^2+i}} \leq \sum_{i=1}^n \frac{1}{\sqrt{n^2+1}} = \frac{n}{\sqrt{n^2+1}}$. In the same way you will find that $a_n = \sum_{i=1}^n \frac{1}{\sqrt{n^2+i}} \geq \sum_{i=1}^n \frac{1}{\sqrt{n^2+n}} = \frac{n}{\sqrt{n^2+n}}$
Therefore your idea is correct and also Kaind's idea to use the sandwich theorem is good. The only thing is that both sequences converge to $1$, not to $0$. For the left one you could use the fact that $\sqrt{n^2+n} \leq n + \sqrt{n}$ and then you get
$$\frac{n}{\sqrt{n^2+n}} \geq \frac{n}{n + \sqrt{n}} = \frac{1}{1+\frac{1}{\sqrt{n}}}$$
and the last fraction obiously converges to $1$.
For the other one we find that
$$\frac{n}{\sqrt{n^2+1}} \leq \frac{n}{\sqrt{n^2}} = 1$$
So with the sandwich theorem we get $a_n \to 1$ as $n \to \infty$.
You're correct in your approach. Just observe that both $\frac{n}{\sqrt{n^2 + n}}$ and $\frac{n}{\sqrt{n^2 + 1}}$ converge to $1$ (substitute $n \rightarrow \frac{1}{x}$) and use sandwich theorem.