Showing that $\sum_{k=1}^\infty\frac{(\log k)^m}{k^{1+\delta}} < \infty $ ? How would you show that $$\sum_{k=1}^\infty\frac{(\log k)^m}{k^{1+\delta}} &lt \infty $$ for $\delta >0$ and $m \in \mathbb{N}$?
 A: Hint:
$$\lim_k \frac{\frac{(\log k)^m}{k^{1+\delta}}}{\frac{1}{k^{1+\frac{\delta}{2}}}}=0$$
and 
$$\sum_{k=1}^\infty\frac{1}{k^{1+\frac{\delta}{2}}} &lt \infty$$
A: 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.
A: 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 $m$:
$$0&lt
\sum_{k=n}^{ n+m} {(\log k)^m\over k^{1+\delta}}\le  
\sum_{k=n}^{ n+m} {k^{\delta\over2}\over k^{1+\delta}}=
\sum_{k=n}^{ n+m} {1\over k^{1+\delta/2}}\ \ 
\buildrel{n, m\rightarrow\infty}\over\longrightarrow\ \ 0;
$$
whence the result follows.
A: 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. 
