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I've been working with the following series:

$$\sum_{n=1}^\infty\frac{\ln^2(n)}{\sqrt{n}(8n+9\sqrt{n})}$$

I know that I must use the comparison test for convergence, but I'm unsure what to compare the series to exactly; if I ignore some things then the series seems to diverge, but if I ignore others the series seems to converge.

Should I used the integral test as an intermediate?

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The series is convergent. Why? First bound it above by a simpler convergent series: $\sum_{n=1}^{\infty} \frac{\text{log}^2{n}}{\sqrt{n}\cdot (8n + 9\sqrt{n})} \leq \sum_{n=1}^{\infty} \frac{\text{log}^2{n}}{8n^{3/2}}$. Now this problem is equivalent to proving that $\sum_{n=1}^{\infty} \text{log}^2(n)\cdot n^{-3/2}$ converges by the limit comparison test. Now why is this series convergent? Try the integral test.

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Hint: Make comparison test with

$$ \frac{\ln^2 n}{ 8 n^{3/2} } .$$

See here for a related problem.

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continuing from chris' series $\sum\limits_{n=1}^{\infty}\frac{ln^2(n)}{8n^{3/2}}$ you can use limit comparison test with the series $$\sum\limits_{n=1}^{\infty}\frac{1}{n^{5/4}}$$ and then the series will converge.

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