# Determine whether the series $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2 +1}$ converges or not.

Determine whether the series $$\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2 +1}$$ converges or not.

** My trial ** I tried dividing $$\frac{\ln(n)}{n^2 +1}$$ by $$1/n^2$$ and finding the limit which was $$\infty$$ so I could not use the limit comparison test and this idea did not work.

Could anyone give me a hint for studying the convergence of this series?

• do you mean $$\sum_{n=1}^{\infty}\frac{\ln n}{n^2+1}$$? – clathratus Jan 16 at 4:52
• Hint: Use the inequality $\ln n < n$ in the following way $\ln n = 2 \ln \sqrt{n} < 2 \sqrt{n}$ – RRL Jan 16 at 4:54
• @clathratus yes sorry I corrected it. – hopefully Jan 16 at 4:56

Compare with $$\sum_{n=1}^{\infty}\frac{1}{n^{1.5}}$$
• Either way. This series converges. With direct comparison, the OP series has smaller terms (eventually). With a limit comparison, the ratio of the OP terms to the terms here converges to $0$. Either implies the OP series converges. – alex.jordan Jan 16 at 5:06
• You would need to know that (eventually) $\ln(n)<n^{0.5}$. There are proofs out there that (eventually) $\ln(n)<n^{\varepsilon}$ for any positive $\varepsilon$. – alex.jordan Jan 16 at 15:41
• For example, $\ln(n)/n^{0.5}$. What is the limit of this as $n\to\infty$? By L'Hospital, it is the same as the limit of $1/(0.5n^{0.5})$, which is $0$. So for large enough $n$, $\ln(n)/n^{0.5}<1$. So for large enough $n$, $\ln(n)<n^{0.5}$. – alex.jordan Jan 16 at 16:39