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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. –  Arturo Magidin Nov 17 '10 at 21:42

5 Answers 5

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? –  Ma.H Nov 17 '10 at 21:34
1  
@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. –  Arturo Magidin Nov 17 '10 at 21:36
    
Got it. Now I see why my approach was incorrect in some of the problems. Thanks! –  Ma.H Nov 17 '10 at 22:01

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). –  Shai Covo Nov 17 '10 at 21:21

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. –  Ma.H Nov 17 '10 at 21:05
    
@Ma.H: I assumed it was a typo, given that you wrote it up so poorly. Please try to format correctly; see meta.math.stackexchange.com/questions/107/… –  Arturo Magidin Nov 17 '10 at 21:12

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.

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