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.

Does this series converge?

$\displaystyle\sum\limits_{n=2}^\infty \frac{\sqrt{n+1}}{(2n^2-3n+1) (\ln n +(\ln n)^2)}$

share|improve this question
What do you think? –  Ron Gordon Jan 16 '13 at 21:58
I tried using the Cauchy Condensation Test, but the resultant expression doesn't seem helpful. –  Ryan Jan 16 '13 at 22:07
The terms of the series are all positive; all you need do is use the Comparison Test, after proving convergence of the series of terms $\sum_{n=2}^\infty \frac{\sqrt{n+1}}{2n^2 - 3n + 1}$ –  Chris Jan 16 '13 at 22:08
Ah, thank you, user_blahblah. –  Ryan Jan 16 '13 at 22:31
@Ryan: is this a problem created by you? It seems unusual for a book. (subjectively speaking) –  Chris's sis the artist Jan 16 '13 at 22:40

4 Answers 4

up vote 3 down vote accepted

Informally, for large $n$, $\sqrt{n+1}$ behaves like $n^{1/2}$, the quadratic in the denominator behaves like $2n^2$, giving a combined behaviour of $\frac{1}{2}\cdot\frac{1}{n^{3/2}}$, plenty good enough for convergence. And the $\log$ stuff at the bottom gives our series a minor (and unnecessary) boost towards convergence.

More formally, we can note that for $n\ge 3$, $$0 \lt \frac{\sqrt{n+1}}{(2n^2-3n+1)(\ln n+\ln^2 n)}\lt \frac{\sqrt{n+1}}{n^2-3n+1}$$ So if we can prove that $\sum_2^\infty \frac{\sqrt{n+1}}{n^2-3n+1}$ converges, it will follow by Comparison that our series converges.

Now note that $\sqrt{n+1}\le 2\sqrt{n}$, and that if $n \ge 6$, then $n^2-3n+1\ge \frac{1}{2}n^2$. It follows that for $n\ge 6$, we have $$\frac{\sqrt{n+1}}{n^2-3n+1}\lt \frac{4}{n^{3/2}}.$$ Since $\sum_2^\infty \frac{1}{n^{3/2}}$ converges, the series $\sum_2^\infty \frac{\sqrt{n+1}}{n^2-3n+1}$ converges.

share|improve this answer
Very nice pedagogical answer, thank you for the insights. –  Ryan Jan 16 '13 at 22:34

There is $n_0$ such that $$\displaystyle\sum\limits_{n=2}^\infty \frac{\sqrt{n+1}}{(2n^2-3n+1) (\ln n +(\ln n)^2)}<\displaystyle\sum\limits_{n=n_0}^\infty \frac{\sqrt{n+1}}{(n+1)\cdot(\sqrt{n+1})^2}=\sum\limits_{n=n_0+1}^\infty \frac{1}{n\cdot\sqrt{n}}$$ Thus the series clearly converges.

share|improve this answer

This one requires no tricks - look at the ratio of the numerator and denominator in the limit $n \to \infty.$ In fact, you can even throw out the $\ln n$ term.

share|improve this answer


share|improve this answer
Please provide your reasoning. –  Ryan Jan 16 '13 at 22:08
@Ryan One might well ask the same of you.... –  Chris Jan 16 '13 at 22:10
@JonasMeyer: this was a really fast answer! (+1):-) –  Chris's sis the artist Jan 16 '13 at 22:18

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.