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I tried to demonstrate the convergence of the following series by root and ratio test but they failed.

$$\sum_{n=1}^{\infty}\frac{\log(n)}{n^2}$$

Then I used Wolfram Alpha, which answered "By the comparison test, the series converges".

How can I do it?

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    $\begingroup$ you should learn that for any $\epsilon > 0$ and $C > 0$, then if $n$ is large enough : $\log(n) < C n^\epsilon$ (this is because $e^x$ is larger than any polynomial when $x \to \infty$) $\endgroup$ – reuns Mar 7 '16 at 23:28
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Prove that $\log(x) \leq \sqrt{x}$ for all $x > 0$. Then you have that $\sum_n \log(n) / n^2 \leq \sum_n 1 / n^{3/2}$. The latter is convergent by the integral criterion since $\int_1^\infty x^{-p}$ converges for all $p > 1$.

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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

for $x>0$. Using the identity $\log(x^{\alpha})=\alpha \log(x)$, we find from $(1)$ that for any $\alpha>0$

$$\frac{n^{-2}-n^{-(2+\alpha)}}{\alpha}\le \frac{\log(n)}{n^2}\le \frac{n^{\alpha-2}-n^{-2}}{\alpha} \tag 2$$

Note that $(2)$ is true for any $\alpha >0$. In particular, it is true for all $0<\alpha <1$.

Thus, we see that - by the comparison test (or the integral test) for example - since the series $\sum_{n=1}^\infty \frac{1}{n^{p}}$ converges for any $1<p$, then the series $\sum_{n=1}\frac{\log(n)}{n^2}$ converges.

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  • $\begingroup$ Not clear. It may take a slightly longer way, but it's correct, and does not leave anything "left to the reader" (as would an answer -- albeit correct -- starting with $\ln n < n^a$ for any $a>0$ and $n$ big enough). $\endgroup$ – Clement C. Mar 7 '16 at 23:33
  • $\begingroup$ @Dr.MV I'd prefer Mauricio G Tec's answer because is a shorter way in order to get my aim. $\endgroup$ – Detry Mar 9 '16 at 8:24
  • $\begingroup$ @Dr.MV I did it. $\endgroup$ – Detry Mar 10 '16 at 23:39

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