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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!

Vim
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3 Answers3

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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.

Daniel Fischer
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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 $$ \sum_{n=1}^\infty\frac1{p_n} \le\sum_{n=1}^\infty\frac{n}{p_1+p_2+\dots+p_n} \le2\sum_{n=1}^\infty\frac1{p_n}\tag{6} $$

Community
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robjohn
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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.

Marco Cantarini
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