Vrftjkry

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?

$lim_{ntoinfty} frac {a_{n+1}}{a_n} = 1.$

$lim_{ntoinfty} (a_n)^{frac 1n} = 1.$

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

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

Write $b=1+2ac$ with $cgt0$. Then show that

$${(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^ato0$$

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

$${(ln n)^aover n^b}le{Bover n^{1+p}}$$

for some $pgt0$, in which case the convergence of $sum{1over n^{1+p}}$ guarantees the convergence of $sum{(ln n)^aover n^b}$. So to be explicit, if we write $b=1+2ac$ and $p=ac$ (with $c={b-1over2a}gt0$), then we have

$${(ln n)^a/n^bover1/n^{1+p}}={(ln n)^a/n^{1+2ac}over1/n^{1+ac}}=left(ln nover n^cright)^a=left(ln n^cover cn^cright)^aleleft(1over ce right)^a$$

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

$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(1over ce right)^asum_{n=1}^infty{1over n^{1+ac}}ltinfty$$

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

$$sum_{n=1}^infty{(ln n)^aover n^b}leleft(2aover (b-1)e right)^asum_{n=1}^infty{1over n^{1+(b-1)/2}}ltinfty$$

By Cauchy condensation test the condensed series is

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

which converges for $b>1$.