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

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.

• You can use the integral test – Brandon Jan 17 '14 at 5:24
• @Brandon Okay, but how would I determine what p has to be from the integral test other than just plugging in numbers and guessing? – Logan Jan 17 '14 at 5:27
• @Brandon Okay, so basically the ln(x) disappears/doesn't really matter? Is this because of a comparison test or something else? I do know that 1/x^p converges if p>1. – Logan Jan 17 '14 at 5:36
• I have voted to reopen this question as the "target" question is specifically about the $p>1$ case. The question here is more general. – user1729 Aug 25 '20 at 13:21

## 1 Answer

$\sum_{n=2}^\infty \frac{1}{n(\ln n)^p}$ 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.

• Awesome! Thanks for such a thorough answer! Can you please explain to me why/how the integral changed for when p=1? – Logan Jan 17 '14 at 5:49
• It is written above. If $p=1$ the integral is $\int \frac{1}{x \ln x}dx=\int \frac{1}{\ln x} d \ln x=\ln (\ln x).$ – kmitov Jan 17 '14 at 8:02
• Oh okay I get it now. Thanks so much! – Logan Jan 17 '14 at 16:02