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$x_n=($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?
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 $(\ln n)^p \lt n$.
