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: Here is another approach
Define
$$
b_m=\sum_{k=2^m+1}^{2^{m+1}}\frac1{a_k}\tag1
$$
By assumption, we have the convergence of
$$
\sum_{m=0}^\infty b_m=\sum_{k=2}^\infty\frac1{a_k}\tag2
$$
Next,
$$
\begin{align}
\sum_{k=2^m+1}^{2^{m+1}}a_kk^p\overbrace{\sum_{k=2^m+1}^{2^{m+1}}\frac1{a_k}}^{b_m}
&\ge\left(\sum_{k=2^m+1}^{2^{m+1}}k^{p/2}\right)^2\tag{3a}\\
&\ge\frac1{(p/2+1)^2}\left(\left(2^{m+1}\right)^{p/2+1}-\left(2^m\right)^{p/2+1}\right)^2\tag{3b}\\
&=\underbrace{\left(\frac{2^{p/2+1}-1}{p/2+1}\right)^2}_{c_p}\,\,2^{m(p+2)}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: Cauchy-Schwarz
$\text{(3b)}$: underestimating a sum with an integral
$\text{(3c)}$: factor out $c_p$
Thus,
$$
\begin{align}
\sum_{k=1}^{2^{m+1}}a_kk^p
&\ge\sum_{k=2^m+1}^{2^{m+1}}a_kk^p\tag{4a}\\
&\ge\frac{c_p}{b_m}2^{m(p+2)}\tag{4b}
\end{align}
$$
Explanation:
$\text{(4a)}$: sum over fewer terms is smaller
$\text{(4b)}$: apply $(3)$
Therefore,
$$
\begin{align}
\sum_{n=1}^\infty\frac{n^{p+1}}{\sum\limits_{k=1}^na_kk^p}
&=\frac1{a_1}+\sum_{m=0}^\infty\sum_{n=2^m+1}^{2^{m+1}}\frac{n^{p+1}}{\sum\limits_{k=1}^na_kk^p}\tag{5a}\\
&\le\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\sum_{m=1}^\infty2^m\frac{2^{(m+1)(p+1)}}{\sum\limits_{k=1}^{2^m}a_kk^p}\tag{5b}\\
&\le\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\frac12\sum_{m=1}^\infty\frac{2^{(m+1)(p+2)}}{\frac{c_p}{b_{m-1}}2^{(m-1)(p+2)}}\tag{5c}\\[6pt]
&=\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\frac{2^{2p+3}}{c_p}\sum_{m=1}^\infty b_{m-1}\tag{5d}\\[6pt]
&=\frac1{a_1}+\frac{2^{p+1}}{a_1+a_22^p}+\frac{2^{2p+3}}{c_p}\sum_{k=2}^\infty\frac1{a_k}\tag{5e}
\end{align}
$$
Explanation:
$\text{(5a)}$: break the sum into the intervals $\left[2^m+1,2^{m+1}\right]$
$\text{(5b)}$: break out the $m=0$ ($n=2$) term
$\phantom{\text{(5b):}}$ for each $m$, there are $2^m$ terms in the sum
$\phantom{\text{(5b):}}$ for each term, $2^m\lt n\le2^{m+1}$
$\text{(5c)}$: apply $(4)$
$\text{(5d)}$: simplify
$\text{(5e)}$: apply $(2)$
A: 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,
$$
\sum_{n=1}^{\infty}\frac{n^{p+1}}{\sum_{k=1}^n a_k k^p}\leq \frac{(p+3)^2}2 \sum_{k=1}^{\infty} \frac1{a_k}. 
$$
A: 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*}
