studying the convergence of a series:

Question

I would like to study the convergence of the series:

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

I could compare the generic element $\frac{\log n}{n^2}$ with $\frac{1}{n^2}$ and say that $$\frac{1}{n^2}<\frac{\log n}{n^2}$$ and $\frac{1}{n^2}$ converges but nothing more about.

Answer 1

HINT

Let use Limit comparison test test with $$\frac{1}{n^{\frac32}}$$

Related OP with a more general discussion Determine if $\sum_{k=1}^{\infty}\frac{\ln{k}}{k\sqrt{k}}$ converges/diverges.

Answer 2

By Cauchy's condensation test, your series converges if and only if$$\sum_{n=1}^\infty\frac{2^n\log2^n}{2^{2n}}$$converges. But this series is equal to$$\sum_{n=1}^\infty\frac{n\log2}{2^n}$$which clearly converges.

Answer 3

Hint Apply the integral test, and integrate by parts.

Answer 4

Here is one way to show it directly. Note that $$\lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\log(n)} = \infty$$ (if you need convinced, just apply L'Hospital's Rule).

Thus $\exists N \in \mathbb{N}, \text{ such that }\sqrt{n} > \log(n) \quad \forall n > N. \quad$ So we write: $$ \begin{align*} \sum_{n=1}^{\infty} \frac{\log(n)}{n^2} = \sum_{n=1}^{N} \frac{\log(n)}{n^2} + \sum_{n=N + 1}^{\infty} \frac{\log(n)}{n^2} &< C_N + \sum_{n=N+1}^\infty \frac{\sqrt{n}}{n^2} \\ &< C_N + \sum_{n=1}^\infty \frac{1}{n^{3/2}} = K \in \mathbb{R^+}\end{align*} $$ Where $C_N > 0$ is some positive constant since $\sum_{n=1}^N \frac{\log(n)}{n^2}$ is a finite, positive sum and $\sum \frac{1}{n^{3/2}}$ converges as it is a $p$-series with $p = \frac{3}{2} > 1$.

Answer 5

Repeat after me: $\ln n\to \infty$ more slowly than any positive power of $n.$ In other words,

$$\frac{\ln n}{n^p} \to 0\,\text { for any } p> 0.$$

Once you have absorbed this, you'll know such things as

$$\frac{\ln n}{n^2}< \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}}$$

for large $n.$ Which in the case of your series proves convergence by the comparison test.