Convergence of sum of $\sum_{n=1}^\infty \frac{\ln n}{n^{2p}}$

Question

I would like to say that this series converges for $p>0.5$ I know that I can use this inequality to compare

$$\frac{1}{n^{2p}}<\frac{\ln n}{n^{2p}}<\frac{n}{n^{2p}}=\frac{1}{n^{2p-1}}$$

where $n\ge 1$ and certainly summing the middle terms, it will diverge if $p\le 0.5$ But the right side, we have convergence if $2p-1>1$ or when $p>1$. So what happens in $(0.5,1)$? What am I not doing?

Answer 1

We have:

1. If $p\in(1,\infty)$, then $\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}<\infty$.

2. If $p\in[0,1],$ then $\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}=\infty$.

Proof: 1. Let $p\in(1,\infty)$. Let $a=\frac{p-1}{2}>0$. Note that when $n$ is sufficiently large, $\ln n<n^{a}$. (For, by L'Hospital rule, $\lim_{x\rightarrow+\infty}\frac{x^{a}}{\ln x}=\lim_{x\rightarrow+\infty}\frac{ax^{a-1}}{\frac{1}{x}}=\lim_{x\rightarrow+\infty}ax^{a}=+\infty$.) Choose $N\in\mathbb{N}$ such that $\ln n<n^{a}$ whenever $n\geq N$. Then $\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}=\sum_{n=1}^{N}\frac{\ln n}{n^{p}}+\sum_{n=N+1}^{\infty}\frac{\ln n}{n^{p}}$. Note that $\sum_{n=N+1}^{\infty}\frac{\ln n}{n^{p}}\leq\sum_{n=N+1}^{\infty}\frac{1}{n^{p-a}}<\infty$ because $p-a=\frac{p+1}{2}>1$ (Power Test). It follows that $\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}<\infty$.

2. Let $p\in[0,1]$, then $\sum_{n=1}^{\infty}\frac{\ln n}{n^{p}}\geq\sum_{n=1}^{\infty}\frac{\ln n}{n}=\infty$ by Integral Test.

Answer 2

if $p\in (0.5,1)$ you can write $p=c+d, c>0.5, d>0$, $\sum {1\over {n^{2c}}}$ converges ${ln(n)\over{n^{2p}}}\over{1\over {n^{2c}}}$=${{ln(n)}\over {n^{2d}}}$ which converges towards $0$ so the comparison test implies that the serie converges.

Answer 3

Note that

$$ \frac{\frac{\ln n}{n^{2p}}}{\frac 1 {n^a}}=\frac{\ln n}{n^{2p-a}}\to0 \implies 2p-a>0\implies a<2p$$

moreover the given series converges by limit comparison test if and only if $a>1$ that is for $2p>1 \implies p>\frac12$.

Answer 4

Here are my 2 cents.

We will show the convergence using Cauchy condensation test, namely:

Lemma: Let $\{a_n\}$ be a positive monotone decreasing sequence. Then we have that \begin{align} \sum^\infty_{n=1} a_n <\infty \ \ \ \text{ if and only if } \ \ \ \sum^\infty_{k=1}2^ka_{2^k}<\infty. \end{align}

Using this fact, we see that \begin{align} \sum^\infty_{k=1}2^ka_{2^k} = \sum^\infty_{k=1}2^k \frac{k\log 2}{2^{2pk}} = (\log 2)\sum^\infty_{k=1} k\left(\frac{1}{2^{2p-1}}\right)^k \end{align} which is summable provided \begin{align} \frac{1}{2^{2p-1}}<1 \ \ \ \Longleftrightarrow \ \ \ 0<(2p-1)\log 2 \ \ \ \Longleftrightarrow \frac{1}{2}<p. \end{align} Hence we see that \begin{align} \sum^\infty_{n=1} \frac{\log n}{n^{2p}}<\infty \ \ \ \text{ if and only if } \ \ \ \frac{1}{2}<p. \end{align}