convergence/divergence $\sum_{n=1}^{\infty} \frac{\ln(n)}{n^2}$

Question

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

I have tried the comparison test with $\frac{1}{n}$ and got $0$ with $\frac{1}{n^2}$ I got $\infty$

What should I try?

Answer 1

Fix $0 < \alpha<1 $. If $n $ is sufficiently large, say $n\geq N$, holds $$\log\left(n\right)\leq n^{\alpha}\tag{1} $$ hence $$\sum_{n\geq1}\frac{\log\left(n\right)}{n^{2}}\leq C + \sum_{n\geq N}\frac{1}{n^{2-\alpha}}<\infty.$$ Maybe it's interesting to note that, if $s>1 $ the Riemann zeta function is defined as $$\zeta\left(s\right)=\sum_{n\geq1}\frac{1}{n^{s}} $$ so in our case we have $$-\frac{d\zeta}{ds}\left(2\right)=\sum_{n\geq1}\frac{\log\left(n\right)}{n^{2}}.$$For prove $(1)$ note that $$\log\left(n\right)\leq n^{\alpha}\Leftrightarrow\frac{\log\left(\log\left(n\right)\right)}{\log\left(n\right)}\leq\alpha $$ and since $\frac{\log\left(\log\left(n\right)\right)}{\log\left(n\right)}\rightarrow0 $ the inequality holds for a sufficiently large $n$.

Answer 2

Since $ x \mapsto \dfrac{\ln x}{x^2}$ is decreasing over $[1,+\infty)$, one may use the integral test : $$ \sum_{n=1}^N\frac{\log\left(n\right)}{n^{2}}\leq \frac{\log\left(N\right)}{N^{2}}+\int_1^N\frac{\log\left(t\right)}{t^2}dt, \quad N\geq1, $$ and observe that, as $N \to \infty$, $$ 0<\int_1^N\frac{\log\left(t\right)}{t^2}dt=1-\frac{1+\log N}N \to 1. $$

Answer 3

Like others suggest, estimating $\ln(n)$ is probably the best way to visualize it.

An alternative is using integral test(and if you get this problem from first year calculus course, this is probably what you're expected to do),

Not hard to check $f(x)=\frac{\ln(x)}{x^2}$ is continuous and decreasing(by computing $f'(x)$).

The integral

$$\int_{1}^{\infty} \frac{\ln(x)}{x^2} dx=$$

Let $u=\ln(x)$, $dv=\frac{1}{x^2}dx$, $du=\frac{1}{x} dx$, $v=-\frac{1}{x}$

$$\int_{1}^{\infty} \frac{\ln(x)}{x^2} dx\\=\frac{-\ln(x)}{x}\Big|_{1}^{\infty}+\int_{1}^{\infty}\frac{1}{x^2} dx\\=0+\frac{-1}{x}\Big|_{1}^{\infty}\\=1$$ converges therefore the original series converge.

Answer 4

We first claim that there exists a positive integer $N$ such that $$\ln(n)\leq n^{1/2},\quad\forall n>N.$$ By L'Hopital's Rule, $$\lim_{n\rightarrow\infty}\frac{\ln(n)}{n^{1/2}}= \lim_{n\rightarrow\infty}\frac{1/n}{\frac{1}{2}n^{-1/2}}= 2\lim_{n\rightarrow\infty}\frac{1}{n^{1/2}}=0.$$ It follows that there exists a positive integer $N$, such that for every $n>N$, $$\frac{\ln(n)}{n^{1/2}}\leq 1\quad\mbox{or}\quad \ln(n)\leq n^{1/2},$$ as desired. Therefore $$\sum_{n=1}^\infty\frac{\ln(n)}{n^2}= \sum_{n=1}^N\frac{\ln(n)}{n^2}+\sum_{n=N+1}^\infty\frac{\ln(n)}{n^2} \leq \sum_{n=1}^N\frac{\ln(n)}{n^2}+\sum_{n=N+1}^\infty\frac{1}{n^{3/2}},$$ where the right series converges. Hence $\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$ converges.