# Series convergence ques

I was just solving another question regarding this topic and just wanna check if I am doing it right.

So the question is whether the given series converges or not, and justify.

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

I have tried using ratio and comparison test, but that leads me nowhere. For comparison test I compared the term with $$\frac{1}{n^2}$$. Am I doing that wrong?

Thank you so much :)

• Compare with $\sum \frac 1 {n^{3/2}}$. – Kavi Rama Murthy Apr 12 '19 at 0:42
• To spell out one item: it is true that $\frac{(\ln n)^2}{n^2} \ge \frac1{n^2}$ for all $n\ge3$; so if the series $\sum_{n=1}^\infty \frac1{n^2}$ diverged, then so would $\sum_{n=1}^\infty \frac{(\ln n)^2}{n^2}$ by the comparison test. However, $\sum_{n=1}^\infty \frac1{n^2}$ does not diverge, so the comparison test doesn't help (it "goes the wrong way"). – Greg Martin Apr 12 '19 at 1:03

$$\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}< \sum_{n=1}^{\infty}\frac{ln(n) \sqrt{n}}{n^2}<\sum_{n=1}^{\infty}\frac{ln(n)}{n^{\frac{3}{2}}}<\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$$

This is a p-series with p>1 so it converges. Thus $$\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}$$ converges as well.

• $\frac{\ln(n)}{n^{\frac{3}{2}}}$ isn't less than $n^{-\frac{3}{2}}$, though. – Ryan Goulden Apr 12 '19 at 1:47

Recall that $$f(s) = \sum_{n=1}^{\infty} n^{-s}$$ converges when $$s>1$$.

Then we observe that $$f''(s)= \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^s}$$, with the same region of convergence.

Hence $$f''(2) = \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^2}$$.

Here's a nice way to show that, for any $$a > 0$$, $$\dfrac{\ln(x)}{x^a} \to 0$$ as $$x \to \infty$$. It actually shows that $$\dfrac{\ln(x)}{x^a} \lt\dfrac{2}{ax^{a/2}}$$ for $$x \gt 1$$.

$$\begin{array}\\ \ln(x) &=\int_1^x \dfrac{dt}{t}\\ &<\int_1^x \dfrac{dt}{t^{1-c}} \qquad\text{since } t^{1-c} < t \text{ for }0 < c \\ &=\int_1^x t^{c-1}dt\\ &=\dfrac{t^c}{c}|_1^x\\ &=\dfrac{x^c-1}{c}\\ &<\dfrac{x^c}{c}\\ \end{array}$$

Let $$c = \frac{a}{2}$$, so that $$\ln(x) \lt \dfrac{x^{a/2}}{a/2} = \dfrac{2x^{a/2}}{a}$$. Dividing by $$x^a$$, $$\dfrac{\ln(x)}{x^a} \lt\dfrac{2}{ax^{a/2}}$$.