# Another series convergence question

Does this series converge?

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

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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) – user 1618033 Jan 16 '13 at 22:40

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.

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

Yes.${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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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.
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.