# Showing that $\sum \frac{\log n}{n^x}$ converges for $x>1$

I'm trying to show that $\sum \frac{\log n}{n^x}$ converges for $x>1$ by the ratio test. Here's what I've got so far $$\frac{a_{n+1}}{a_n} = \frac{\log (n+1) n^x}{(n+1)^x \log n}$$ $$=\left(\frac{n}{n+1}\right)^x \frac{\log (n+1)}{\log n}$$ but I can't see how to manipulate the $\frac{\log (n+1)}{\log n}$ term to make this congerge to a limit less than 1, can anyone help?

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L'Hopital will show $\lim\limits_{n\rightarrow\infty}{\log(n+1)\over\log n}$ is 1. But the limit of the other expression is 1 also. So the Ratio test will fail. – David Mitra Apr 12 '12 at 13:32
Hint: $\frac{logn}{n^x}=o(\frac{1}{n^{\frac{1+x}{2}}})$ – 89085731 Apr 12 '12 at 13:39

The reason you can't do that is because the limit of those terms is $1,$ so the ratio test won't work here. Instead, note that $\displaystyle f(t) = \frac{ \log t}{t^x}$ is eventually decreasing, since $\displaystyle f'(t) = \frac{1- x \log t}{t^{x+1}}$. Then either the integral test or Cauchy condensation will finish this off.

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For $x>1$, let $x=1+2a, a>0$, then $$\lim_{n\to\infty}\frac{\frac{\log n}{n^{1+2a}}}{\frac{1}{n^{1+a}}}=\lim_{n\to\infty}\frac{\log n}{n^{a}}=0.$$

Because the series $\sum\frac{1}{n^{a+1}}$ converges, so does $\sum \frac{\log n}{n^{x}}.$

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I don't know if this is right, but what I did was I decided to say that (ignoring $n=1$)

$\log n = n^t$ or $t=\log(\log n)/\log(n)$, and say that $x=1+f$ so then I could rewrite the sum as $n^t/n^x$ or, $n^{t-x}$ or, $n^{t-f-1}$ and since $t$ approaches zero as $n$ becomes large, $f$ will eventually exceed $t$. The sum of the terms before that point will be finite, and the sum of the terms after that point will be convergent. Finite plus finite equals finite. (this is likely wrong)

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