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

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$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

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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

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

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