# Test for convergence of $\sum_{n=2}^\infty(\ln n)^p$

What test do you suggest for testing the convergence of the series $\sum_{n=2}^\infty(\ln n)^p$ when $p<0$? I have (I hope I did it correctly) already tested the convergence for $p\ge0$ using the ratio test in this way $$\limsup_{n\to\infty}\left|\left(\frac{\ln (n+1)}{\ln n}\right)^p\right|=+\infty\ge1.\qquad(1)$$ The limit superior is greater or equal than 1 and for this reason the series diverges when $p\ge0$.

Edit

I was wrong in (1). If $p>0$ the series diverges because $(\ln n)^p$ does not converge to zero. The question remains the same. How do I study the case $p<0$?

• What makes you think this limit is $\infty$? – J.R. Jan 24 '14 at 13:40
• Yes you're right, it does not tend to infinity! I will edit the question. – Charlie Jan 24 '14 at 13:45
• @TooOld So how can I do to test the convergence when $p<0$? Can you give me a hint? – Charlie Jan 24 '14 at 14:15
• See the answer below by barto. – J.R. Jan 24 '14 at 14:28

An alternative would be by showing that $\displaystyle\sum_{n>2}(\ln n)^p\gg\sum_{n>2}\frac1n$.
To see why, note that $$\lim_{n\to+\infty}\frac{n}{(\ln n)^{-p}}\overset{(t=\ln n)}{=}\lim_{t\to+\infty}\frac{e^t}{t^{-p}}=+\infty,$$ which implies $n\cdot(\ln n)^p>1$ for $n$ large enough.
Since $\displaystyle\sum_{n>2}\frac1n$ diverges, so does $\displaystyle\sum_{n>2}(\ln n)^p$.
• Always remember: $\ln x$ is nothing compared to $x$, which is again nothing compared to $e^x$. – punctured dusk Jan 24 '14 at 13:56
• Interestingly, $\sum\limits_{k=2}^{\infty} \ln(k)^{-k}$ does nonetheless converge (to something around $3.2426$). I wonder what the closed form is... – G. H. Faust Jan 24 '14 at 14:23
• @Charlie, it's by comparison. He shows that eventually, the individual terms of the series become larger than the corresponding terms in the divergent series $\sum {1\over{n}}$. – G. H. Faust Jan 24 '14 at 14:26
• @Charlie: What I essentially do is comparing the series $\sum(\ln n)^{p}$ to the harmonic series $\sum\frac1n$. It is well known that the harmonic series diverges. If we can show that every term of the series $\sum(\ln n)^{p}$ is bigger that the corresponding term $\frac1n$ in the harmonic series, from a certain index, then certainly the series $\sum(\ln n)^{p}$ will diverge too. – punctured dusk Jan 24 '14 at 14:26
• To show that $(\ln n)^{p}>\frac1n$ for all $n>\delta$, I use a certain limit. Since that particular limit is $+\infty$, we know that $\frac n{(\ln n)^{-p}}>1(=\epsilon)$ for every $n>\delta$. Hence $(\ln n)^{p}>\frac1n$ if $n>\delta$. – punctured dusk Jan 24 '14 at 14:30