How can I check the convergence of the sequence? Does it diverge?

How can I check the convergence of the sequence $\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}$? I think that it diverges,because it is bounded below from $\frac{n(n+1)}{2\sqrt{n^2+n}}$ and above from $\frac{n(n+1)}{2\sqrt{n^2+1}}$..Is this correct?

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Try to use integral test. – Mhenni Benghorbal Feb 11 '14 at 19:30
Yes, it's correct. Now you could work a little with your bounds to make the $\sim \frac{n}{2}$ behaviour more obvious. – Daniel Fischer Feb 11 '14 at 19:32
@DanielFischer So then can I use the Squeeze Theorem and find that the limit is $+ \infty$ ?? – Mary Star Feb 11 '14 at 19:34
Not sure whether the Squeeze theorem covers that situation in the formulation you have, but if you relax your bounds a bit you get $$\frac{n}{2} < a_n < \frac{n+1}{2}.$$ – Daniel Fischer Feb 11 '14 at 19:38
Call your sequence $(a_n)$. You have observed (lower bound) that $a_n\gt \frac{\sqrt{n(n+1)}}{2}$. That is enough to show divergence to $\infty$. You do not need your (correct) upper bound. Where both bounds would be useful is in showing that $\lim_{n\to\infty}\frac{a_n}{n}=\frac{1}{2}$. – André Nicolas Feb 11 '14 at 19:54

HINT: you can use Squeeze Theorem.

$\frac{1}{\sqrt{n^2+n}}+\frac{2}{\sqrt{n^2+n}}+\cdots+\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+1}}+\cdots+\frac{n}{\sqrt{n^2+1}}$.

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So are the bounds that I have found wrong? – Mary Star Feb 11 '14 at 19:32
Nope you just found different bounds. – mne__povezlo Feb 11 '14 at 19:34
Your bounds are tighter, @MaryStar. – Daniel Fischer Feb 11 '14 at 19:35
The lower bound here is not strong enough, and we do not need an upper bound. – André Nicolas Feb 11 '14 at 19:39
@Babgen: I hope you don't mind. I edited the right hand side – robjohn Feb 11 '14 at 21:52

Since $n^2\le n^2+k\le\left(n+\frac12\right)^2$ for $0\le k\le n$, we have $$\frac1{n+\frac12}\sum_{k=1}^nk\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac1n\sum_{k=1}^nk$$ Thus, $$\frac n2\le\frac{n(n+1)}{2n+1}\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac{n+1}2{}$$ That is, the sequence diverges.

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Following my comment. The sum can be estimated as

$$\sum_{k=1}^{n} \frac{k}{\sqrt{n^2+k} } \sim \int_{1}^{n} \frac{x}{\sqrt{n^2+x} } dx=\dots\,.$$

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