Check whether function series is convergent (uniformly):
$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n}\ln \left( \frac{x}{n} \right)$ for $x\in[1;+\infty)$
I don't know how to do that.
Check whether function series is convergent (uniformly):
$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n}\ln \left( \frac{x}{n} \right)$ for $x\in[1;+\infty)$
I don't know how to do that.
The series doesn't converge. Use integral test or Cauchy-condensation-test via monotonicity of the general term
Fix an $x$ and think about this tail of the series $\sum_{n=3x}^{\infty}\frac{1}{n}\ln{\left(\frac{x}{n}\right)} = -\sum_{n=3x}^{\infty}\frac{1}{n}\ln{\left(\frac{n}{x}\right)} \leq -\sum_{n=3x}^{\infty}\frac{1}{n}\ln{3} \leq -\sum_{n=3x}^{\infty}\frac{1}{n} \leq 0$ Now use what you know about $\sum_{n=1}^{\infty}\frac{1}{n}$ to conclude.