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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.

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up vote 0 down vote accepted

The series doesn't converge. Use integral test or Cauchy-condensation-test via monotonicity of the general term

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in integral test or Cauchy condensation test general term must be non negative and monotone decreasing so assumptions are not satisfied.. – ray Jun 13 '12 at 20:41
The general term may well be not positive: put a minus sign in front of the series. As for the monotonicity, note that $$ \left(\frac 1t\ln \frac xt\right)' =\frac{\ln t}{t^2} -\frac{1+\ln x}{t^2}$$ so is asymptotically incresing. The assumption are satisfied except at most for a finite number of terms but this doesn't affect the convergence – Unoqualunque Jun 14 '12 at 9:43

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.

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