How to determine does this series converge or diverge? I have tried d'Alembert's ratio test but in the limit I get $1$. I suppose I should compare it with some other series, but I can't figure out with which one to compare to.
$$\sum\limits_{n=1}^{\infty}\dfrac{\ln(\sqrt{n})}{n}$$
Answer
It diverges:
$\sum\limits_{n=1}^{\infty}\dfrac{\ln(\sqrt{n})}{n} =
\sum\limits_{n=1}^{\infty}\dfrac{\ln(n^{\frac{1}{2}})}{n} =
\sum\limits_{n=1}^{\infty}\dfrac{\frac{1}{2}\ln(n)}{n} =
\frac{1}{2}\sum\limits_{n=1}^{\infty}\dfrac{\ln(n)}{n}\geq
\frac{1}{2}\sum\limits_{n=1}^{\infty}\dfrac{1}{n} =
\infty$
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