$Is the infinite series $\sum_{n=1}^{\infty}\frac{1}{n \ln n}$ convergent or divergent?$

Question

$Is the infinite series $\sum_{n=1}^{\infty}\frac{1}{n \ln n}$ convergent or divergent?$

Is the infinite series $\sum_{n=1}^{\infty}\frac{1}{n \ln n}$ convergent or divergent?

Which convergence test should be utilized? The ratio and comparision tests don't work.

Calculus Precalculus Algebra

Rosetta

59

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer.

Answer 1

Answers can be viewed only if

The questioner was satisfied and accepted the answer, or
The answer was disputed, but the judge evaluated it as 100% correct.

View the answer

I guess the lower bound of the sume should be $n=2$ and not $n=1$.

We can use the integral test to determine the convergence of this series. By the integral test, this series converges if and only if the integral

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

Then $du=\frac{1}{x}dx$. So
$$\int \frac{1}{x\ln x}dx=\int \frac{1}{u}du=\ln|u|+c=\ln (|\ln x|)+c$$.

Thus $\int_{2}^{\infty}\frac{1}{x \ln x}dx=\ln (|\ln x|)|_2^{\infty}=\lim_{x \rightarrow \infty} \ln (|\ln x|)-\ln (|\ln 2|)=\infty$, (Note that that $\lim_{x \rightarrow \infty} \ln (x)=\infty$). Hence the integral $$\int_{2}^{\infty}\frac{1}{x \ln x}dx$$ diverges, and hence the infinite sum in your question also diverges.

Erdos

4.6K

$Is the infinite series $\sum_{n=1}^{\infty}\frac{1}{n \ln n}$ convergent or divergent?$

Answer

Related Questions

Search