Find the values of $p$ for which $\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ is convergent

Question

Find the values of $p$ for which the series $\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ is convergent.

I know that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent if $p>1$ and divergent if $p\leq1$.

So with the series in question, I can't just say "when $p>1$" right? Because the denominator is $n(\ln n)^p$ not just $n^p$ as the formula states. So how should I go about tackling this problem? I know that doing a u-sub for $\ln n$ would work because $du$ would equals $\frac{dn}{n}$ but what would I do with the $p$?

Thanks for the help.

Answer 1

$\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ You can use the integral test

$\int_{2}^\infty \frac{1}{x(\ln x)^p}dx=\lim_{A \to \infty}\int_{2}^A\frac{1}{x(\ln x)^p}dx=\lim_{A \to \infty}\int_2^A (\ln x)^{-p}d (\ln x)$

Let $p>1$ then $\lim_{A \to \infty}\int_2^A (\ln x)^{-p}d (\ln x)$ $=\lim_{A\to \infty}\left[\frac{(\ln A)^{1-p}}{1-p}-\frac{(\ln 2)^{1-p}}{1-p}\right]$ $=-\frac{(\ln 2)^{1-p}}{1-p}$ because $1-p<0$ and $\ln A \to \infty$ as $ A \to \infty$.

Therefore in this case integral and seris are convergent.

Let $p<1$ then $\lim_{A \to \infty}\int_2^A (\ln x)^{-p}d (\ln x)$ $=\lim_{A\to \infty}\left[\frac{(\ln A)^{1-p}}{1-p}-\frac{(\ln 2)^{1-p}}{1-p}\right]$ $=\infty$ because $1-p>0$ and $\ln A \to \infty$ as $ A \to \infty$.

Therefore in this case integral and seris are divergent.

Let $p=1$ then $\lim_{A \to \infty}\int_2^A (\ln x)^{-1}d (\ln x)$ $=\lim_{A\to \infty}\left[\ln(\ln A)-\ln(\ln 2)\right]$ $=\infty$ because $\ln A \to \infty$ as $ A \to \infty$.

Therefore in this case integral and seris are divergent.