convergence of $\sum^\infty_{k=1} \frac {\ln(k)}{k^p}$

Question

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?

One can do better. There is convergence if $p\gt 1$ and divergence otherwise. — André Nicolas, Jul 12 '15 at 02:13
It is not convergent for $p=1$, the comment said we have convergence for $p\gt 1$ and divergence otherwise. The otherwise includes $p=1$. — André Nicolas, Jul 12 '15 at 02:19
@TanMath: the derivative of $\zeta$ function IS a function too... so it representation as a series converges, i.e., if a series converges it represent a function (if Im not wrong). — Masacroso, Jul 12 '15 at 02:51
@Masacroso but I must use the four series tests: ratio, root, limit comparison, and comparison tests — TanMath, Jul 12 '15 at 02:53
I know @TanMath... I was trying to explain the comment of Lucian... What I said is so advanced for you surely — Masacroso, Jul 12 '15 at 02:55
Possible duplicate of Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ — Nosrati, Nov 03 '18 at 06:25

André Nicolas · Accepted Answer · 2015-07-12T18:57:09.477

2

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.

edited Jul 12 '15 at 18:57

answered Jul 12 '15 at 02:17

André Nicolas

507,029

Can you please add more steps? – TanMath Jul 12 '15 at 02:24
Also, I don't see how $\lim_{k\to\infty} \frac{\ln k}{k^{d/2}}=0$ connects with $\sum_1^\infty \frac{1}{k^{1+d/2}}$ – TanMath Jul 12 '15 at 02:27
1

I will, if they are still needed, but only quite a bit later. Two reasons: (i) you might as well have a try and (ii) I am about to go out to eat. – André Nicolas Jul 12 '15 at 02:27
About your second comment, in the limit comparison we will be using $\frac{\ln k}{k^{1+d}}=\frac{1}{k^{1+d/2}}\cdot \frac{\log k}{k^{d/2}}$. – André Nicolas Jul 12 '15 at 02:30
"We do some of the details of showing that the series converges if $p\gt 0$" do you mean $p\gt1$? – TanMath Jul 12 '15 at 18:48
I certainly do mean $p\gt 1$! Thanks for spotting it. – André Nicolas Jul 12 '15 at 18:50
and is it 1+(d/2), not (1+d)/2? – TanMath Jul 13 '15 at 04:53
It is $1+(d/2)$, which is the usual meaning of $1+d/2$. I decided not to write $k^{1+\frac{d}{2}}$ because that sometimes becomes illegible on small screens. – André Nicolas Jul 13 '15 at 04:57

Eugene Zhang · Answer 2 · 2015-07-12T19:37:52.827

1

By L'Hospital's Rule, for any $\epsilon>0$ $$ \lim_{x\to\infty}\frac{\ln{x}}{x^{\epsilon}}=\lim_{x\to\infty}\frac{1}{\epsilon x^{\epsilon}}=0 $$ So $\ln{x}=o(x^{\epsilon})$.

For any $p>1$, let $\epsilon=\dfrac{p-1}{2}$. So $$ \epsilon>0\hspace{ 5mm} \text{and} \hspace{ 5mm} p-\epsilon=\dfrac{p+1}{2}>1 $$ So $$ \sum_{n=1}^{\infty}\frac{\ln{n}}{n^p}=\sum_{n=1}^{\infty}\frac{\ln{n}}{n^{\epsilon}n^{p-\epsilon}}<\sum_{n=1}^{\infty}\frac{1}{n^{p-\epsilon}}<\infty $$ For $p\leqslant 1$ $$ \sum_{n=1}^{\infty}\frac{\ln{n}}{n^p}>\sum_{n=1}^{\infty}\frac{1}{n}=\infty $$

edited Jul 12 '15 at 19:37

answered Jul 12 '15 at 03:32

Eugene Zhang

16,805

line 4, how do you know there is an $\epsilon > 0$, I cant understand taht – Jul 12 '15 at 05:36
I fixed it. Thanks. – Eugene Zhang Jul 12 '15 at 05:56
You don't need L'Hospital's rule to know $\ln x=o(x^\varepsilon)$ for all $\varepsilon$: this is a basic fact in Asymptotic Analysis. – Bernard Jul 12 '15 at 10:47
@Bernard, I just prove it in an easy way. $\ln{x}=o(x^ε)$ is not obvious. Of course there are other ways to do it. – Eugene Zhang Jul 12 '15 at 19:39

convergence of $\sum^\infty_{k=1} \frac {\ln(k)}{k^p}$

2 Answers2