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I've reached my wit's end with this problem. We have not been taught about Ratio Tests or anything involving Radius of Convergence, but we have a problem asking for it.

For what values of $p$ does the sum converge? $$\sum_{n=1}^\infty{\ln(n) \over{n^p}}$$

I notice that it is "similar" to

$$\sum_{n=1}^\infty{\frac {1}{n^p}}$$

And we know from the Integral Test of Convergence that this series converges when $\lvert p\rvert > 1$. Does this have any connection to the answer? Because I also notice that (via WolframAlpha) that when $p=1$ the series diverges, but when $p=1.1$ it converges.

user319742
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    The Ratio Test is not useful for this problem. The series $\sum \frac{1}{n^p}$ is very relevant. You will also need the fact that $\frac{\ln n}{n^t}$ has limit $0$ for any positive $t$. It will turn out that we have divergence if $p\le 1$ and convergence if $p\gt 1$. – André Nicolas Mar 04 '16 at 00:59

3 Answers3

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Note that $\ln n<n$ so

$$\sum_{n=1}^\infty \frac{\ln n}{n^p}<\sum_{n=1}^\infty\frac{n}{n^p}=\sum_{n=1}^\infty \frac{1}{n^{p-1}}$$

which is the familiar $p$-series. Thus the series converges by the comparison test if $p-1>1$, or if $p>2$.

This is not a full proof since it remains to show whether the series converges for $1< p\le 2$. Note that divergence for $p\le1$ is given by the same comparison test above.

pancini
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  • someone want to explain the down vote? – pancini Mar 06 '16 at 22:46
  • It wasn't me. I still think a down-vote should require a justification or be moderated. This can be done without compromising anonymity. – DanielWainfleet Jul 06 '18 at 19:20
  • Hard to say, since down voting is an important mechanic. It is mostly annoying when someone downvotes and you can't figure out why, but its fairly rare. – pancini Jul 06 '18 at 19:22
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For any $d>0$ we have $\lim_{x\to \infty} (\ln x)/x^d=0.$

For $p>1$ let $d=(p-1)/2.$ Then for all but finitely many $n$ we have $0<(\ln n)/x^d<1.$ So for all but finitely many $n$ we have $$0<(\ln n)/x^p<x^d/x^p=1/ x^{p-d}=1/x^{(p+1)/2}.$$ And $(p+1)/2>1.$ So by the Comparison test, $\sum_n (\ln n)/n^p$ converges.

For $p\leq 1$ we have $$n\geq 3\to (\ln n)/n^p>1/n^p >1/n>0,$$ so again by Comparison, $\sum_n (\ln n)/n^p$ diverges.

DanielWainfleet
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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$

for $x>0$. Then, for any number $\alpha >0$, we have from $(1)$

$$\frac{x^{\alpha}-1}{x^{\alpha}}\le \log(x^\alpha)\le x^{\alpha}-1 \tag 2$$

Using $\log(x^{\alpha})=\alpha \log(x)$ in $(2)$ reveals that for any $x>0$ and any $\alpha >0$

$$\frac{x^{\alpha}-1}{\alpha x^{\alpha}}\le \log(x)\le \frac{x^{\alpha}-1}{\alpha} \tag 3$$

Finally, using $(3)$ we find that

$$\frac{n^{-p}-n^{-(\alpha+p)}}{\alpha }\le \frac{\log(n)}{n^p}\le \frac{n^{\alpha-p}-n^{-p}}{\alpha}\tag 4$$

  • From the right-hand side of $(4)$, for any $p>1$, we can choose $\alpha$ so that $p-\alpha >1$ also. Therefore, if $p>1$, then we find that

$$\sum_{n=1}^N \frac{\log(n)}{n^p}\le \frac{1}{\alpha}\sum_{n=1}^N\frac{1}{n^{p-\alpha}} \tag 5$$

and the series of interest converges by the comparison test (or the integral test).

  • From the left-hand side of $(4)$, for any $p\le 1$ and for all $\alpha$, we see that

$$\sum_{n=1}^N \frac{\log(n)}{n^p}\ge \frac{1-2^{-\alpha}}{\alpha}\sum_{n=2}^N \frac{1}{n^p}$$

and the series of interest diverges by the comparison test (or the integral test).

Community
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Mark Viola
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