# Showing that $\sum_{k=1}^\infty\frac{(\log k)^m}{k^{1+\delta}} < \infty$ ?

How would you show that $$\sum_{k=1}^\infty\frac{(\log k)^m}{k^{1+\delta}} < \infty$$ for $\delta >0$ and $m \in \mathbb{N}$?

-
Maybe show that after a while $(\log k)^m<k^{\delta/2}$. – André Nicolas Apr 23 '12 at 16:30

Hint:

$$\lim_k \frac{\frac{(\log k)^m}{k^{1+\delta}}}{\frac{1}{k^{1+\frac{\delta}{2}}}}=0$$

and

$$\sum_{k=1}^\infty\frac{1}{k^{1+\frac{\delta}{2}}} < \infty$$

-

I'd probably use Cauchy' condensation test. In this case, you have to prove the convergence of the series $$\sum_{n} \frac{(n \log 2)^m}{2^{\delta n}}\simeq \sum_n \frac{n^m}{2^{\delta n}}.$$ The last series converges, since the exponential growth of the denominator suffices to kill the polynomial growth at the numerator. Actually, the root test works, too.

-
I believe the appropriate summand is $(n\log 2)^m/2^{\delta n}$. – user26872 Jun 25 '12 at 0:41
Thank you, I had forgotten the term $2^n$ in front of everything. – Siminore Jun 25 '12 at 9:54
Glad to help. I like this approach. (+1) – user26872 Jun 25 '12 at 16:22

Using L'Hôpital's rule repeatedly, and being somewhat lazy:\eqalign{ \lim_{k\rightarrow\infty}{k^{\delta/2}\over (\log k)^m} &=\lim_{k\rightarrow\infty} {{\delta\over 2}k^{{\delta\over2}-1} \over m(\log k)^{m-1}\cdot{1\over k}} \cr &=C_1\lim_{k\rightarrow\infty}{k^{\delta\over 2}\over(\log k)^{m-1} }\cr &=C_2\lim_{k\rightarrow\infty}{k^{\delta\over 2}\over(\log k)^{m-2} }\cr &\ \vdots\cr &\ \cr &=C_m\lim_{k\rightarrow\infty}{k^{\delta\over 2} },\cr } for some positive constants $C_1$, $\ldots\,$, $C_m$. Since $\lim\limits_{k\rightarrow\infty}{k^{\delta\over 2} }=\infty$, it follows that $\lim\limits_{k\rightarrow\infty}{k^{\delta/2}\over (\log k)^m}=\infty.$ Consequently, there is an $N$ so that for $k\ge N$, we have $(\log k)^m\le k^{\delta\over2}$.

Then, for any $n\ge N$ and nonnegative integer $m$: $$0< \sum_{k=n}^{ n+m} {(\log k)^m\over k^{1+\delta}}\le \sum_{k=n}^{ n+m} {k^{\delta\over2}\over k^{1+\delta}}= \sum_{k=n}^{ n+m} {1\over k^{1+\delta/2}}\ \ \buildrel{n, m\rightarrow\infty}\over\longrightarrow\ \ 0;$$ whence the result follows.

-

Let $a_k = (\log k)^m/k^{1+\delta}$. In the limit, the ratio of successive terms goes like $$\frac{a_{k+1}}{a_k} = 1 - \frac{1+\delta}{k}.$$ Since $1+\delta > 1$ the series converges. This is essentially Raabe's test.

-