Yesterday I was tutoring a high school student on the convergence of infinite series. We encounter the following series on his practice assignment:
$$\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}$$
He's at the precalculus level so he hasn't seen the integral test which would be the standard method for determining that this particular series diverges.
His teacher though gave him a hint that the formula for infinite interest compound was to be used. I assumed that we had to use something along the lines of:
$$e^x=\lim_{n\to\infty}\left( 1+\frac{x}{n}\right)^n\implies x=\lim_{n\to\infty} n \ln \left(1+\frac{x}{n}\right)$$
I couldn't do it in the end, it takes me to weird places. Is there really a way of determining this series is divergent without using the integral test?
Edit: The tests that the teacher taught my student and thus the only ones allowed are the following:
nth term test, Comparison/limit comparison test, alternating series test, ratio test.
Best Answer
To use the hint you had: notice that $$\ln\ln(n+1) - \ln\ln n = \ln\frac{\ln(n+1)}{\ln n} = \ln\left( 1+ \frac{\ln(1+1/n)}{\ln n} \right)$$ Now, the series $$ \sum_{n=2}^N (\ln\ln(n+1) - \ln\ln n) $$ is telescopic and diverges. And by the limit comparison test, the two series $$ \sum_n \ln\left( 1+ \frac{\ln(1+1/n)}{\ln n}\right), \qquad \sum_n \frac{1}{n\ln n} $$ have same behavior since $\lim_{n\to\infty}\frac{\ln\left( 1+ \frac{\ln(1+1/n)}{\ln n}\right)}{\frac{1}{n\ln n}} = 1$.