For what values of p does $\sum^\infty_{k=1} \frac {\ln(k)}{k^p}$
converge?
Here is my work:
$\ln (k) < k$ on $[1,\infty)$ so $\frac {\ln (k)}{k^p} < \frac {k}{k^p}$
Therefore,
$ \sum^\infty_{k=1} \frac {k}{k^p} = \sum^\infty_{k=1} \frac {1}{k^{p-1}} $
This converges for $p>2$ so the original sum must converge for $p>2$ as well. Now, how do I find when this infinite sum diverges?
Best Answer
We have divergence if $p\le 1$, by comparison with $\sum_1^\infty \frac{1}{k}$.
We have convergence if $p\gt 1$. One way to do it is to let $p=1+d$, and do a limit comparison with $\sum_1^\infty \frac{1}{k^{1+d/2}}$. The key fact is that $\lim_{k\to\infty} \frac{\ln k}{k^{d/2}}=0$.
Added: We do some of the details of showing that the series converges if $p\gt 1$. Call our series $\sum_1^\infty a_k$. Let $p=1+d$. Let $b_k=\frac{1}{k^{1+d/2}}$.
The series $\sum_{k=1}^\infty b_k$ converges. We show that $\sum_{k=1}^\infty a_k$ converges by Limit Comparison with $\sum_{k=1}^\infty b_k$. We have $$\frac{a_k}{b_k}=\frac{\ln k}{k^{1+d}}\cdot k^{1+d/2}=\frac{\ln k}{k^{d/2}}.$$ But by L'Hospital's Rule, or otherwise, $\lim_{x\to\infty}\frac{\ln x}{x^{d/2}}=0$, so by Limit Comparison $\sum_{k=1}^\infty a_k$ converges.
More informally, the $\frac{1}{k^{d/2}}$ part of $\frac{1}{k^{1+d}}$ was used to "kill" the $\ln k$ (in the long run). What remains, namely $\frac{1}{k^{1+d/2}}$, decays fast enough to ensure convergence.
If you prefer, you can prove the convergence when $p\gt 1$ by using the Cauchy Condensation Test.