2
$\begingroup$

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.

$$\sum\limits_{n=1}^{\infty}\dfrac{\ln(\sqrt{n})}{n}$$

$\endgroup$
1
  • 3
    $\begingroup$ Compare with $\sum \frac{1}{n}$. $\endgroup$
    – OR.
    Mar 28, 2014 at 9:30

2 Answers 2

4
$\begingroup$

It diverges:

$\sum\limits_{n=1}^{\infty}\dfrac{\ln(\sqrt{n})}{n} = \sum\limits_{n=1}^{\infty}\dfrac{\ln(n^{\frac{1}{2}})}{n} = \sum\limits_{n=1}^{\infty}\dfrac{\frac{1}{2}\ln(n)}{n} = \frac{1}{2}\sum\limits_{n=1}^{\infty}\dfrac{\ln(n)}{n}\geq \frac{1}{2}\sum\limits_{n=1}^{\infty}\dfrac{1}{n} = \infty$

$\endgroup$
4
$\begingroup$

Hint

$$\frac{\ln\left(\sqrt n\right)}{n}\ge \frac1n\quad\text{for n large enough}$$

$\endgroup$
2
  • $\begingroup$ To me that isn't obvious right away. How could I prove that if I had to use it in a test? $\endgroup$
    – Nick
    Mar 28, 2014 at 9:40
  • $\begingroup$ $\sqrt{n} \rightarrow +\infty$ therefore when you compose by $\ln$ it goes to infinity as well. Alternatively, $ln(\sqrt{n})>1 \Leftrightarrow n > \exp{2}$ $\endgroup$
    – T_O
    Mar 28, 2014 at 9:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .