Infinite sum: prove or disprove this statement In one of my textbook I was asked to prove:

Suppose $0<p_1<p_2<\cdots<p_n<\cdots$, prove:
  $$\sum_{n=1}^{\infty}\frac{1}{p_n}\quad\text{converges}\Leftrightarrow\sum_{n=1}^{\infty}\frac{n}{p_1+p_2+\cdots+p_n}\quad\text{converges}$$

My notion was that I should be able to prove that when $n\to\infty$, 
$$\frac{1}{p_n}\sim\frac{n}{p_1+p_2+\cdots+p_n}$$
but I failed.
I could not even think of an effective method to prove the $\implies$ part, my failed attempt is as follows:
If $\sum_{n=1}^{\infty}\frac{1}{p_n}$ converges, then I want to use comparison test: 
$$\frac{\frac{n}{p_1+p_2+\cdots+p_n}}{\frac{1}{p_n}}\le\frac{\sum_{n=1}^{\infty}\frac{1}{p_n}}{n\cdot\frac{1}{p_n}}$$
I tried to upper-bound RHS, but since $\sum_{n=1}^{\infty}\frac{1}{p_n}$ converges, by comparison test we have
$$\frac{\frac{1}{p_n}}{\frac1n}\to 0^+\quad\text{as}\quad n\to\infty$$
and thus I could not bound RHS.
Can anyone help me with this problem? Best regards!
 A: 
My notion was that I should be able to prove that when $n\to\infty$, 
  $$\frac{1}{p_n}\sim\frac{n}{p_1+p_2+\cdots+p_n}$$
  but I failed.

That's not surprising, since the asymptotic equality doesn't hold in general. If for example $p_n = 2^n$, then
$$\frac{n}{p_1 + \dotsc + p_n} = \frac{n}{2^{n+1}-1} \sim \frac{n}{2p_n}.$$
But, by monotonicity, we have
$$\sum_{i=1}^n p_i < np_n,$$
and thus the majorisation
$$\frac{1}{p_n} < \frac{n}{p_1 + \dotsc + p_n}$$
shows the one direction of the equivalence. On the other hand, for $n \geqslant 2$, we have
$$\sum_{i=1}^n p_i \geqslant \sum_{i = \lfloor n/2\rfloor}^n p_i \geqslant \frac{n}{2}p_{\lfloor n/2\rfloor}$$
and hence
$$\frac{n}{p_1 + \dotsc + p_n} < \frac{2}{p_{\lfloor n/2\rfloor}},$$
which shows the other direction.
A: Since
$$
\frac1{p_n}\le\frac{n}{p_1+p_2+\dots+p_n}\tag{1}
$$
if
$$
\sum_{n=1}^\infty\frac{n}{p_1+p_2+\dots+p_n}\tag{2}
$$
converges, then
$$
\sum_{n=1}^\infty\frac1{p_n}\tag{3}
$$
converges.

For the other direction, we can use the same argument as in this answer.
By Cauchy-Schwarz, we have
$$
\begin{align}
\left(\sum_{j=1}^kp_j\right)\left(\sum_{j=1}^k\frac{j^2}{p_j}\right)
&\ge\left(\sum_{j=1}^kj\right)^2\\[3pt]
&=\frac{k^2(k+1)^2}{4}\tag{4}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^kp_j}
&\le\sum_{k=1}^n\frac4{k(k+1)^2}\sum_{j=1}^k\frac{j^2}{p_j}\\
&=\sum_{j=1}^n\frac{j^2}{p_j}\sum_{k=j}^n\frac4{k(k+1)^2}\\
&\le\sum_{j=1}^n\frac{j^2}{p_j}\sum_{k=j}^n2\left(\frac1{k^2}-\frac1{(k+1)^2}\right)\\
&\le\sum_{j=1}^n\frac{j^2}{p_j}\frac2{j^2}\\
&=2\sum_{j=1}^n\frac1{p_j}\tag{5}
\end{align}
$$
Therefore, if $(3)$ converges, $(2)$ converges. In fact, we have
$$
\sum_{n=1}^\infty\frac1{p_n}
\le\sum_{n=1}^\infty\frac{n}{p_1+p_2+\dots+p_n}
\le2\sum_{n=1}^\infty\frac1{p_n}\tag{6}
$$
A: By partial summation we have $$\sum_{k\leq n}p_{k}=np_{n}-\sum_{k\leq n-1}k\left(p_{k+1}-p_{k}\right)
 $$ then we can write the second series as $$\sum_{n\geq1}\frac{n}{np_{n}-\sum_{k\leq n-1}k\left(p_{k+1}-p_{k}\right)}.
 $$ Consider $$\frac{np_{n}-\sum_{k\leq n-1}k\left(p_{k+1}-p_{k}\right)}{np_{n}}=1-\frac{\sum_{k\leq n-1}k\left(p_{k+1}-p_{k}\right)}{np_{n}}.
 $$ We have $$0\leftarrow1-\frac{\left(n-1\right)\left(p_{n}-p_{1}\right)}{np_{n}}<1-\frac{\sum_{k\leq n-1}k\left(p_{k+1}-p_{k}\right)}{np_{n}}<1-\frac{p_{n}-p_{1}}{np_{n}}\rightarrow1
 $$ and so $$\lim_{n\rightarrow\infty}1-\frac{\sum_{k\leq n-1}k\left(p_{k+1}-p_{k}\right)}{np_{n}}=c\in\left(0,1\right)
 $$ and so by limit comparison test if one of these two series converges, the other series converges.
