Can I find the limit of $\sum_\limits{i=1}^{n^2}\frac{1}{\sqrt{n+i}}$ with the squeeze theorem? Problem: Calculate limit of $\frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \frac{1}{\sqrt{n+3}} + \cdots + \frac{1}{\sqrt{n+n^2}}$ as $n$ approaches infinity.
Solution: Denote the above some as $X$, then we can bound it:
$$ \infty\longleftarrow\frac{1}{\sqrt{n+n^2}} \lt X \lt \frac{n^2}{\sqrt{n+1}} \lt \frac{n^2}{\sqrt{n}} = \sqrt{\frac{n^4}{n}}\longrightarrow \infty.$$
So, from the Squeeze Principle, $\lim X = \infty$. Am I doing the right thing?
 A: In proving something diverges to $\infty$, you don't need to squeeze it both from below and from above: it is enough to "half-squeeze" it from below by something that goes to $\infty$ (because the limit you want is getting "pushed up" to $\infty$). So here, you can just note that each summand is greater than or equal to $\frac{1}{\sqrt{n+n^2}}$, so the sum of $n^2$ of them is at least as large as $\frac{n^2}{\sqrt{n+n^2}}$, and since $\lim\limits_{n\to\infty}\frac{n^2}{\sqrt{n+n^2}}=\infty$ and $\frac{n^2}{n+n^2}\leq X(n)$ for each $n$ (notice that what you called $X$ actually varies with each $n$, so you should not give it a name that makes it look like it is fixed), then $\lim\limits_{n\to\infty}X(n)=\infty$ as well.
A: Yes, you are doing the right thing, assuming you meant $n^2 / \sqrt{n+n^2} < X$ rather than $1 / \sqrt{n+n^2} < X$.
A: First  $\frac{1}{\sqrt{n+n^{2}}} \rightarrow 0 $ if $n\rightarrow \infty $.
Futhermore, in the infinite case the squeeze principle used so:
if $a_{n}\leq b_{n}$ and $a_{n}\rightarrow \infty$ then $b_{n}\rightarrow \infty$.
Also you sequence i do not understand, 
A: The term $\frac{1}{\sqrt{n+n^2}}$ goes to zero as $n \rightarrow \infty$, so that side doesn't work.  You should think about how many terms you are adding.
A: Each term is bounded from below by the last term of your sum. Therefore a lower bound for the sum is $n^2$ times the last term. Show that this lower bound diverges as $n$ goes to infinity.
In this case, you only really need a lower bound. Don't bother with an upper bound.
