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I'm trying to solve the problem :

$$\sum_{n=1}^{\infty}\frac{\ln(n)}{n^x}$$ is the above series convergent in $x\in(\frac{3}{2},\infty)$?

I found that in $(2,\infty)$ it is convergent, but I don't know how to prove it for the rest of the interval. $$\sum_{n=1}^{\infty}\frac{\ln(n)}{n^x} \le \sum_{n=1}^{\infty}\frac{n}{n^x} = \sum_{n=1}^{\infty}\frac{1}{n^{x-1}}$$ So for $x > 2$ it's convergent.

How should I proceed?

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  • $\begingroup$ Hint: $\ln(n)<\sqrt n$. Actually, $\ln(n)<(n^\alpha-1)/\alpha$ for all $\alpha>0$, giving a stronger result... $\endgroup$ – user65203 Jun 16 '17 at 19:02
  • $\begingroup$ Yeah, i just noticed that! Thanks!! $\endgroup$ – Zarrie Jun 16 '17 at 19:04
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Hint. one may observe there exists $n_0>1$ such that for all $n\ge n_0$, $$ \left|\frac{\ln n}{n^{1/4}}\right|\le 1 $$ giving $$ \left|\frac{\ln n}{n^{x}}\right|\le \frac{1}{n^{x-1/4}},\qquad n\ge n_0, $$ leading to a convergent series for $x-\dfrac 14>1$ that is for $x>\dfrac 54$ the latter being true for $x>\dfrac32$.

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    $\begingroup$ That really helped me! Actually i proved it observing $lim_{n} \frac{ln n}{n^1/\alpha}$ Thanks! $\endgroup$ – Zarrie Jun 16 '17 at 19:05
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For $t\ge1$, you have $\dfrac1t\le1$ with equality holding at $t=1$. Then by integration from $1$ to $t$,

$$\ln t\le t-1.$$

And substituting $t$ with $t^\alpha$ where $\alpha>0$,

$$\ln t\le\frac{t^\alpha-1}\alpha.$$

This yields,

$$\sum_{n=1}^\infty\frac{\ln n}{n^\beta}\le\sum_{n=1}^\infty\frac{n^\alpha-1}{\alpha n^\beta}=\sum_{n=1}^\infty\frac{1}{\alpha n^{\beta-\alpha}}-\sum_{n=1}^\infty\frac{1}{\alpha n^\beta}.$$

So for any $\beta\in(1,\infty)$ you will find an $\alpha>0$ such that all these series converge.


This result generalizes to

$$\sum_{n=1}^\infty\frac{\ln^\gamma n}{n^\beta}.$$

The logarithm grows slower than any power.

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