Test for convergence $\sum_{n = 2}^\infty \frac{1}{(n+1)\ln^2(n+1)}$

Question

Test for convergence $$\sum_{n = 2}^\infty \frac{1}{(n+1)\ln^2(n+1)}$$

Here's my attempt! I decided to use the integral test for this.

$$\frac{1}{(n+1)\ln^2(n+1)} > \frac{1}{(n+1)^2\ln^2(n+1)}$$

Set $f(x) = \dfrac{1}{x\ln(x)}$ and let $x = (n + 1)^2$

$$\int_9^\infty\frac{1}{x\ln(x)}dx = \int_{\ln(9)}^\infty\frac{du}{u} = \ln(u)_{\ln(9)}^\infty = \infty$$

In this case, does $\ln^2(x)$ mean $(\ln x)^2$ or $\ln\ln x$? ${}\qquad{}$ — Michael Hardy, Jul 09 '14 at 19:59
http://math.stackexchange.com/questions/655240/convergence-of-series-involving-in-iterated-logarithms-sum-frac1n-log-n — Lost1, Jul 09 '14 at 20:16
Why don't you just let your series start at 3 and replace n+1 with n? This question has been asked like 1 billion times here. — Lost1, Jul 09 '14 at 20:17

score 2 · Answer 1 · answered Jul 09 '14 at 20:01

You can use Integral test for convergence for

$$f(x)=\frac{1}{x\ln^2x}$$

$f$ is monotone decreasing on $[1,\infty)$, so series converges to a real number if and only if the improper integral:

$$\int_{1}^{\infty} f(x)dx$$

is finite.

But this integral is finite, because:

$$\int \frac{1}{x\ln^2x}dx=-\frac{1}{\ln x}$$

score 1 · Answer 2 · answered Jul 09 '14 at 20:01

1

Since $\frac{1}{k \log^2 k}$ is a strictly decreasing function for $k \geq 2$, we can use the integral test and compare it to $\int_{1}^{\infty} \frac{dx}{x \log^2 x}$ to see if it converges.

answered Jul 09 '14 at 20:01

Alex

19,262

Test for convergence $\sum_{n = 2}^\infty \frac{1}{(n+1)\ln^2(n+1)}$

2 Answers2