Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent. 
Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and
$p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is
convergent.

Since
$$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}=\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p},$$
then
\begin{align*} \frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \sqrt[k]{(k!)^p}}\sim \frac{k^{p+1}}{\sqrt[k]{a_1a_2\cdots a_k}\cdot \frac{k^p}{e^p}}= e^p\cdot \frac{k}{\sqrt[k]{a_1a_2\cdots a_k}}. \end{align*}
This perhaps can not work.

This is similar to robjohn's solution and it is based on the solution of this page: http://www.math.org.cn/forum.php?mod=viewthread&tid=28918
By Cauchy-Schwarz, we have
$$
\sum_{k=1}^n a_k k^p \sum_{k=1}^n \frac k{a_k} \geq \left(\sum_{k=1}^n k^{\frac{p+1}2}\right)^2\geq \left( \frac{n^{\frac{p+3}2}}{\frac{p+3}2}\right)^2=\frac{4n^{p+3}}{(p+3)^2}.
$$
Then
$$
\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}4 \frac1{n^2}\sum_{k=1}^n \frac k{a_k}.
$$
Summing over $n$, we have
$$
\sum_{n=1}^{\infty}\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}4 \sum_{n=1}^{\infty}\frac1{n^2}\sum_{k=1}^n \frac k{a_k}.
$$
Interchanging order of the summation on the right side, we have
$$
\sum_{n=1}^{\infty}\frac1{n^2}\sum_{k=1}^n \frac k{a_k}=\sum_{k=1}^{\infty} \frac k{a_k} \sum_{n=k}^{\infty} \frac1{n^2}\leq \sum_{k=1}^{\infty} \frac k{a_k} \left( \frac1{k^2} + \frac1k\right)\leq \sum_{k=1}^{\infty} \frac 2{a_k}.
$$
Hence,
$$
\sum_{n=1}^{\infty}\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}2 \sum_{k=1}^{\infty} \frac1{a_k}. 
$$
