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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?

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  • $\begingroup$ do you mean $$\sum_{n=1}^{\infty}\frac{\ln n}{n^2+1}$$? $\endgroup$ – clathratus Jan 16 at 4:52
  • $\begingroup$ Hint: Use the inequality $\ln n < n$ in the following way $\ln n = 2 \ln \sqrt{n} < 2 \sqrt{n}$ $\endgroup$ – RRL Jan 16 at 4:54
  • $\begingroup$ @clathratus yes sorry I corrected it. $\endgroup$ – hopefully Jan 16 at 4:56
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Compare with $$\sum_{n=1}^{\infty}\frac{1}{n^{1.5}}$$

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  • $\begingroup$ you mean comparison not limit comparison? $\endgroup$ – hopefully Jan 16 at 4:56
  • $\begingroup$ 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. $\endgroup$ – alex.jordan Jan 16 at 5:06
  • $\begingroup$ but if I use the direct comparison test how can I compare them? $\endgroup$ – hopefully Jan 16 at 14:19
  • $\begingroup$ 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 $\varepsilon$. $\endgroup$ – alex.jordan Jan 16 at 15:41
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    $\begingroup$ 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 $n$, $\ln(n)<n^{0.5}$. $\endgroup$ – alex.jordan Jan 16 at 16:39

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