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?

  • $\begingroup$ 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

Compare with $$\sum_{n=1}^{\infty}\frac{1}{n^{1.5}}$$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.