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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?

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    I had assumed the first fraction in the inequality was a typo when I TeX-ed up the question; since several responders commented, I've restored it (after TeX-ing) to its original statement.2010-11-17

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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.

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    Might be a trivial question, but why isn't it correct to lower bound with just the most smallest element (I mean, instead of ${n^2}$)? Isn't that inequality correct?2010-11-17
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    @Ma.H: Yes, you can bound below by just the smallest summand (or by *any* summand, for that matter, since they are all positive). But being able to bound is not the same as being able to reach the conclusion you want. Here, each summand $\frac{1}{\sqrt{n+k}}$ does *not* go to $\infty$ as $n\to \infty$, it goes to $0$. So out of *that* bound the only thing you get is that the limit of $X(n)$ is greater than or equal to $0$; which, while true, is *not* what you want to conclude.2010-11-17
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    Got it. Now I see why my approach was incorrect in some of the problems. Thanks!2010-11-17
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Yes, you are doing the right thing, assuming you meant $n^2 / \sqrt{n+n^2} < X$ rather than $1 / \sqrt{n+n^2} < X$.

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    This answer corresponds to the original post (cf. Ross' answer).2010-11-17
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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,

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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.

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    Yes, my bad on that. It goes zero indeed. So, my bounding was incorrect in this case.2010-11-17
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    @Ma.H: I assumed it was a typo, given that you wrote it up so poorly. Please try to format correctly; see http://meta.math.stackexchange.com/questions/107/what-should-go-in-the-math-stackexchange-faq/255#2552010-11-17
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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.