How to tell if $\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$ converges using Integral Test?

Question

So I have this problem:

Determine whether or not the following infinite series converge or diverge. State what test you used.

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

So I decided to do the Integral Test for $$\int^\infty_1\frac{ln(n)}{n^2}dx$$

I made $u = \ln(n)$ and $du = \frac{1}{n}dx$ and got:

$$\int^\infty_1\frac{u}{n}*du$$

which would become:

$$u[\ln(n)]|^\infty_1$$

This is incomplete, but does this look right so far? If not, how do I fix it?

When you tried to integrate after using a $u$ substitution., you left both $u$'s and $n$'s in the expression and chose to ignore one (and even integrated with respect to the wrong term). This should not be done. — JMoravitz, Sep 13 '18 at 18:44
You should find if you are more careful that the integral is equal to exactly $1$. Try integrating by parts. — JMoravitz, Sep 13 '18 at 18:48
Recall that when you make a change of variables in the integrand, you also have to change the limits of integration. — saulspatz, Sep 13 '18 at 18:48
Possible duplicate of Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ — Nosrati, Nov 03 '18 at 06:44

user · Accepted Answer · 2018-09-13T19:01:28.417

By integral test we should obtain

$$\int_1^\infty \frac{\ln x}{x^2} dx=\left[-\frac{1+\log x}{x}\right]_1^\infty$$

or by $\ln x=u \implies \frac1x dx=du$

$$\int_0^\infty \frac{u}{e^u} du=\left[-\frac{u+1}{e^u}\right]_0^\infty$$

If you are not forced to use integral test, as an effective alternative, we can use limit comparison test with

$$\sum_{n=1}^\infty\frac{1}{n^p}$$

with $p>1$ such that

$$\frac{\frac{ln(n)}{n^2}}{\frac{1}{n^p}}\to 0$$

score -1 · Answer 2 · answered Sep 13 '18 at 19:35

Since

$\begin{array}\\ \ln(n) &=\int_1^n \frac{dt}{t}\\ &<\int_1^n \frac{dt}{t^{1/2}}\\ &=\int_1^n t^{-1/2}dt\\ &=2t^{1/2}|_1^n\\ &<2n^{1/2}\\ \end{array} $,

$\dfrac{\ln(n)}{n^2} \lt \dfrac{2n^{1/2}}{n^2} = 2\dfrac1{n^{3/2}} $ and the sum of this converges by the integral test.

This can be modified to show that $\sum \frac{\ln(n)}{n^{1+c}} $ converges for any $c > 0$. Of course the sum diverges for $c = 0$.

score -1 · Answer 3 · answered Sep 14 '18 at 07:40

Denote $$x_n=\frac{\ln n}{n^2},~~~y_n=\frac{1}{n^{3/2}}.$$We have $$\lim_{n \to \infty}\frac{x_n}{y_n}=\lim_{n \to \infty}\frac{\ln n}{n^{1/2}}=\lim_{n \to \infty}\frac{\dfrac{1}{n}}{\dfrac{1}{2}n^{-1/2}}=2\lim_{n \to \infty}\frac{1}{n^{1/2}}=0.$$ Notice that $\sum y_n$ is convergent. By the comparison test of the limit form, $\sum x_n$ is also convergent.

How to tell if $\sum_{n=1}^\infty\frac{\ln(n)}{n^2}$ converges using Integral Test?

3 Answers3