I want to test for this series Convergence: $$\sum^{\infty}_{n=2}{\frac{1}{\sqrt n}\ln\left(\frac{n+1}{n-1}\right)}$$ I'm wondering which method should i use?

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    $\begingroup$ Hint: for $x$ near $1$, we have $\ln(x) \sim x -1$ so for large $n$, $\ln\left( \frac{n+1}{n-1}\right) \sim \frac 2{n-1}$. $\endgroup$ – User8128 Jun 16 '17 at 4:51
  • $\begingroup$ @User8128 can you explain to me how can i use this hint to test for convergence? $\endgroup$ – Tel0s Jun 16 '17 at 5:02
  • $\begingroup$ @Arjang actually it converges $\endgroup$ – User8128 Jun 16 '17 at 5:03
  • $\begingroup$ @Arjang yes but i need to proof, how can you see it? $\endgroup$ – Tel0s Jun 16 '17 at 5:03
  • $\begingroup$ @Tel0s using my above comment, the summand is asymptotic to $1/n^{3/2}$ so the sum coverages by the comparison test/$p$-series test. $\endgroup$ – User8128 Jun 16 '17 at 5:06

You can use the comparison test.

Note that for $x > 0$ we have that $\ln(x) \leq x-1$. Hence $\ln (\frac{n+1}{n-1}) \leq \frac{n+1}{n-1} -1 = \frac{2}{n-1}$.

Hence, $\frac{1}{\sqrt{n}} \ln(\frac{n+1}{n-1}) \leq \frac{2}{\sqrt{n}(n-1)}$ so the series converges by the comparison test and the $p$-test.

  • $\begingroup$ First, you have $\ln(x)\le x-1$ but then you switch to $\ln\left(\frac{n+1}{n-1}\right)\le1-\frac{n+1}{n-1}$. $\endgroup$ – John Wayland Bales Jun 16 '17 at 5:23
  • $\begingroup$ @JohnWaylandBales corrected, thank you. $\endgroup$ – Mustafa Said Jun 16 '17 at 5:33
  • $\begingroup$ You're welcome. $\endgroup$ – John Wayland Bales Jun 16 '17 at 5:33

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