How would you show that \sum_{k=1}^\infty\frac{(\log k)^m}{k^{1+\delta}} < \infty for $\delta >0$ and $m \in \mathbb{N}$?
Showing that \sum_{k=1}^\infty\frac{(\log k)^m}{k^{1+\delta}} < \infty ?
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3Maybe show that after a while (\log k)^m
. – 2012-04-23
4 Answers
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.
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0Glad to help. I like this approach. (+1) – 2012-06-25
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.