Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
Try to use integral test. –  Mhenni Benghorbal Feb 11 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 at 19:32
@DanielFischer So then can I use the Squeeze Theorem and find that the limit is $+ \infty$ ?? –  Mary Star Feb 11 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 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 at 19:54

3 Answers 3

HINT: you can use Squeeze Theorem.


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

share|improve this answer

Following my comment. The sum can be estimated as

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

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.