Converge or diverge? $\sum_{n=1}^{\infty}\frac {(\ln n)^a}{n^b}$

Question

Converge or diverge? $\sum_{n=1}^{\infty}\frac {(\ln n)^a}{n^b}$, $a > 0, b > 1$.

How do you approach this? Since every criteria I know gives $1$. What should I do?

$\lim_{n\to\infty} \frac {a_{n+1}}{a_n} = 1.$

$\lim_{n\to\infty} (a_n)^{\frac 1n} = 1.$

I think this should converge for every $a,b$ with the restriction in hypothesis, right?

Answer 1

Write $b=1+2ac$ with $c\gt0$. Then show that

$${(\ln n)^a/n^{1+2ac}\over1/n^{1+ac}}=\left(\ln n\over n^c\right)^a\to0$$

Remarks (added later): What we're really after is an inequality of the form

$${(\ln n)^a\over n^b}\le{B\over n^{1+p}}$$

for some $p\gt0$, in which case the convergence of $\sum{1\over n^{1+p}}$ guarantees the convergence of $\sum{(\ln n)^a\over n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1\over2a}\gt0$), then we have

$${(\ln n)^a/n^b\over1/n^{1+p}}={(\ln n)^a/n^{1+2ac}\over1/n^{1+ac}}=\left(\ln n\over n^c\right)^a=\left(\ln n^c\over cn^c\right)^a\le\left(1\over ce \right)^a$$

where we use the fact that $f(x)={\ln x\over x}$ attains its maximum at $x=e$, where $f(e)={1\over e}$. It follows that

$$\sum_{n=1}^\infty{(\ln n)^a\over n^b}\le\left(1\over ce \right)^a\sum_{n=1}^\infty{1\over n^{1+ac}}\lt\infty$$

where $c={b-1\over2a}$. Writing it without the $c$, it's

$$\sum_{n=1}^\infty{(\ln n)^a\over n^b}\le\left(2a\over (b-1)e \right)^a\sum_{n=1}^\infty{1\over n^{1+(b-1)/2}}\lt\infty$$

Answer 2

Hint Apply the integral test, that is, consider the convergence of $$\int_1^{\infty} \frac{\log^a x \,dx}{x^b} = \int_0^{\infty} u^a e^{(1 - b) u} du .$$

Answer 3

By Cauchy condensation test the condensed series is

$$\sum_{n=1}^{\infty}\frac {2^n(\ln 2^n)^a}{2^{bn}}=\sum_{n=1}^{\infty}\frac {n^a\ln^a 2}{2^{n(b-1)}}$$

which converges for $b>1$.