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Could you please give me some hint how to decide about convergence of the series $\sum_{n\ge1}\frac{ln\left(\frac{n+1}n\right)}{\sqrt n}$ ?

I tried using comparison test: $\frac{ln\left(\frac{n+1}n\right)}{\sqrt n}\ge \frac {ln\left(\frac1n\right)}{\sqrt n}=-\frac {ln(n)}{\sqrt n}$. Series $\sum_{n\ge1}\frac{ln(n)}{\sqrt n}$ diverges by integral test, but for comparison test all compared sequenced must be non-negative and $ln\left(\frac1n\right)\le0$ for all n.

Thanks.

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    $\begingroup$ You have $ln(1+\frac{1}{n}) \sim \frac{1}{n}$ when $n$ goes to infinity. What can you say according to a serie comparison argument? $\endgroup$ – Pi89 Apr 30 '14 at 13:10
  • $\begingroup$ You should find a better replacement for $\ln \frac{n+1}{n} = \ln \left(1+\frac{1}{n}\right)$. $\endgroup$ – Daniel Fischer Apr 30 '14 at 13:10
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Yes it does using the asymptotic comparison:

$$\frac{\ln\left(\frac{n+1}n\right)}{\sqrt n}\sim_\infty\frac{1}{n\sqrt n}$$

Remark For the comparison test the general term of the series must have an unchanged sign i.e. not alternating series (no matter positive or negative).

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