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Please guys i need help with this limit:

$$\lim_{n \to \infty} \left(\frac {1}{\sqrt{n^2+1}}+ \frac{1}{\sqrt{n^2+2}}+\dots +\frac{1}{\sqrt{n^2+2n}}\right)$$

I don't know what to do?

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I assume that the limit should be taken as $n\to\infty$? –  Brian M. Scott Jan 5 '13 at 18:15
    
use this –  Santosh Linkha Jan 5 '13 at 18:32
    
Thank you, that is what i was searching for. –  aleksamarkoni Jan 5 '13 at 18:35
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2 Answers

HINT

Note that $$\underbrace{\dfrac{2n}{n+1} < \dfrac{2n}{\sqrt{n^2+2n}}}_{n^2 +2n < (n+1)^2} \leq \sum_{k=1}^{2n} \dfrac1{\sqrt{n^2+k}} \leq \underbrace{\dfrac{2n}{\sqrt{n^2+1}} < \dfrac{2n}{\sqrt{n^2}}}_{n^2 < n^2+1} = 2$$

EDIT

Make use of the sandwich theorem to conclude that the limit is $2$.

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Marvis thanks for a quick replay, but can you explain it with a bit more details, i am sorry but i don't know what to do with this form. Sorry for trouble you. Or maybe some link or reference that can be helpful, thanks in advance. –  aleksamarkoni Jan 5 '13 at 18:25
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@aleksamarkoni: Marvis is pointing out that the $n$-th term of your sequence is trapped between $$\frac{2n}{\sqrt{(n+1)^2-1}}$$ and $$\frac{2n}{\sqrt{n^2+1}}\;;$$ what are the limits of these fractions as $n\to\infty$? –  Brian M. Scott Jan 5 '13 at 18:30
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$$\frac{2n}{\sqrt{n^2+2n}}=\frac{2}{\sqrt{1+\frac{2}{n}}}\xrightarrow [n\to\infty]{}2$$

Can you do now something similar with the RHS in Marvis's answer and then use the squeeze theorem?

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Thank you, i did not know about squeeze theorem, so i understand perfectly now. Thank you all very much. –  aleksamarkoni Jan 5 '13 at 18:37
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