Limit of $\sum\limits_{k=1}^{2n}\frac{ 1}{\sqrt{n^2+k}}$

Question

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?

I assume that the limit should be taken as $n\to\infty$? – Brian M. Scott Jan 05 '13 at 18:15 — Brian M. Scott, Jan 05 '13 at 18:15
Thank you, that is what i was searching for. – aleksamarkoni Jan 05 '13 at 18:35 — aleksamarkoni, Jan 05 '13 at 18:35

score 4 · Answer 1 · 2013-01-05T19:42:24.250

4

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 $2$.

edited Jan 05 '13 at 19:42

answered Jan 05 '13 at 18:15

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 05 '13 at 18:25
2

@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 05 '13 at 18:30

score 0 · Answer 2 · answered Jan 05 '13 at 18:32

0

$$\frac{2n}{\sqrt{n^2+2n}}=\frac{2}{\sqrt{1+\frac{2}{n}}}\xrightarrow [n\to\infty]{}2$$

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

answered Jan 05 '13 at 18:32

DonAntonio

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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 05 '13 at 18:37

Limit of $\sum\limits_{k=1}^{2n}\frac{ 1}{\sqrt{n^2+k}}$

2 Answers2