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.

A  solution (in Chinese) is as follows. Is it correct?
\begin{align*}  \sum_{k=1}^n\frac{k^{p+1}}{a_1+2^pa_2+\cdots+k^pa_k}&\le \sum_{k=1}^n\frac{k^{p+1}}{k\sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}}\\ &=\sum_{k=1}^n\left(\frac{k}{\sqrt[k]{k!}}\right)^p\frac{1}{\sqrt[k]{a_1a_2\cdots a_k}}\\& \le e^p\sum_{k=1}^n\frac{1}{\sqrt[k]{a_1a_2\cdots a_k}}\\ &\le e^{p+1}\cdot \sum_{k=1}^n\frac{1}{a_k}.~~~~~~~~\color {blue}{\text{(Carleman's Inequality)}}   \end{align*}
