# Discuss the convergence or the divergence of the series

$x_n=($ln $n)^{-p}$ is the nth term of the series I am working on.

I have tried looking at the series for different ranges of $p$. I also have noted that the ratio test is inconclusive.

My work:

If $p\leq0$, nth term does not go to 0 as n$\rightarrow\infty$. Thus the series diverges.

If $0<p<1$, the series diverges by the limit comparison test with $y_n=\sum\frac{1}{n(lnn)^p}$.

However, I don't know how to tackle the case for when $p\geq1$. Could you give me an idea?

• For large $n$ we have $(\ln n)^p \lt n$. So the series diverges. Jun 8, 2014 at 17:09
• @AndréNicolas Thank you for your comment. But how do you prove the inequality? Jun 8, 2014 at 17:19

Let $p\gt 1$. If we can show that for large enough $n$ we have $(\ln n)^p \lt n$, then Comparison with $\sum \frac{1}{n}$ will show that our series diverges.
Let $w=\ln n$. We want to show that for large $w$, we have $w^p \lt e^w$, or equivalently that $w\lt e^{w/p}$. Note that in fact by L'Hospital's Rule we have $$\lim_{w\to\infty} \frac{w}{e^{w/p}}=0.$$ It follows that if $w$ is large enough, then $w^p\lt e^w$.
If we don't want to use L'Hospital's Rule, we can use the series for $e^x$ to show that for all positive $w$ we have $e^{w/p}\gt 1+\frac{w}{p}+\frac{w^2}{2p^2}\gt \frac{w^2}{2p^2}$. Thus if $w$ is positive then $$\frac{w}{e^{w/p}} \lt \frac{2p^2}{w}.$$ This can be used to find an explicit $N$ such that if $n\ge N$ we have $(\ln n)^p \lt n$.
• You are welcome. It is a good idea to separate the easy $p\le 0$. In what I wrote, I assumed $p\gt 1$, since you had done the rest, but the argument works word for word for any $p\gt 0$. And very importantly, only $n$ large ever matters. The series $\sum_1^\infty a_n$ converges if and only if $\sum_{2014}^\infty a_n$ converges. Jun 8, 2014 at 17:45