Conditions for convergence of $\sum_{n = 1}^{+\infty} \frac{\ln(n)}{n^\alpha}$

Question

I'm trying to find when series $$\sum_{n = 1}^{+\infty} \frac{\ln(n)}{n^\alpha}$$ converges or diverges, but unfortunately without use of integral calculus (as we haven't reached it yet in our real analysis course). My problems book says it is preferable to use p-series ratio test.

For $\alpha = 1$, we can compare series with $\sum_{n = 1}^{+\infty} \frac{1}{n}$, which is divergent, and so is original series then (since n-th term of it is greater than n-th term of harmonic series). For $\alpha < 1$ the case is same.

For $\alpha > 1$ my logic is as follows:

Taking $b_n = \frac{1}{n^{\alpha - \epsilon}}$, $a_n = \frac{\ln n}{n^\alpha}$ and taking $\lim \frac{a_n}{b_n} = \lim \frac{\ln n}{n^\varepsilon}$ we get, that for this series to converge $\alpha - \varepsilon > 1$ and for limit to be finite it is required to $\varepsilon > 0$. This gives us $\alpha > 1 + \varepsilon$. Since $\alpha > 1 + \varepsilon$ it holds that $\alpha > 1$.

Is my attempt correct?

Looks pretty correct. Note that Cauchy condensation test is also applicable here. — Simply Beautiful Art, Apr 02 '17 at 00:05
$\lim_{x\to\infty}\frac{\ln x}{x^\varepsilon} = 0$ for any $\varepsilon > 0$. — Friedrich Philipp, Apr 02 '17 at 00:07
I can't follow your reasoning, as you begin by saying "For a>1" and end by saying "Therefore it holds that a>1" — DanielWainfleet, Apr 02 '17 at 11:27

score 1 · Answer 1 · edited Jun 12 '20 at 10:38

The development in the OP is solid. I thought it might be instructive to present another way forward. To that end, we begin with a primer.

PRIMER:

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

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1<x}\tag{P1}$$

for $x>0$. Now, using $\log(x^b)=b\log(x)$, then we see from $(P1)$ that for $b>0$,

$$\log(x)\le \frac{x^b-1}{b}<\frac{x^b}{b}\tag {P2}$$

Note from $(P2)$, with $x=n$, that $\log(n)\le \frac{n^b-1}{b}<\frac{n^b}{b}$ for all $b>0$. Therefore,

$$\frac{\log(n)}{n^\alpha}<\frac{n^{b}}{bn^\alpha} \tag 1$$

The inequality in $(1)$ is valid for all $b>0$ including those values of $b$ such that $\alpha-b>1$. Therefore, taking $b<a-1$, we see that the series converges by the "$p$ test" for $\alpha>1$.

Now, if $\alpha\ge 1$ the series diverges since we have

$$\frac{\log(n)}{n^\alpha}\ge \frac1n$$

and the series diverges by comparison with the harmonic series.

Putting it all together, we see that the series $\sum_{n=1}^\infty \frac{\log(n)}{n^\alpha}$ converges for all $\alpha >1$ and diverges for all $\alpha\le 1$.

Good answer, I got it instantly. :-) – Accelerate to the Infinity Apr 02 '17 at 14:05 — Accelerate to the Infinity, Apr 02 '17 at 14:05
Thank you. And pleased to hear! -Mark – Mark Viola Apr 02 '17 at 15:07 — Mark Viola, Apr 02 '17 at 15:07

score 1 · Answer 2 · answered Apr 02 '17 at 16:31

The series $$ \zeta(\alpha) = \sum_{n\geq 1}\frac{1}{n^\alpha} $$ is absolutely convergent by the p-test for any $\alpha>1$. Since $\log(n)\ll n^{\varepsilon}$ for any $\varepsilon>0$ and any $n$ large enough, the series $$ \sum_{n\geq 1}\frac{\log(n)}{n^\alpha} $$ is absolutely convergent for any $\alpha>1$ as well, and by termwise differentiation and Euler's product

$$\begin{eqnarray*}\forall \alpha>1,\qquad \color{blue}{\sum_{n\geq 1}\frac{\log(n)}{n^\alpha}} = -\zeta'(\alpha) &=& \zeta(\alpha)\cdot\left(-\frac{d}{d\alpha}\log\zeta(\alpha)\right)\\&=&\left(\color{green}{\sum_{n\geq 1}\frac{1}{n^{\alpha}}}\right)\cdot\left(\color{purple}{\sum_{p\in\mathcal{P}}\frac{\log p}{p^\alpha-1}}\right)\end{eqnarray*} $$

where $\mathcal{P}$ is the set of prime numbers.

Conditions for convergence of $\sum_{n = 1}^{+\infty} \frac{\ln(n)}{n^\alpha}$

2 Answers2