Let $(a_n)$ be a sequence of positive numbers such that $\displaystyle \sum_{n=1}^{\infty} a_n$ is convergent. Show that there is a sequence $(M_n)$ such that $M_n\to\infty$ and $\displaystyle\sum_{n=1}^{\infty}M_n a_n$ converges.
I tried examples by taking $a_n$ to be terms of the harmonic or geometric series and could find a desired $M_n$. I’m not sure how to do it in general.
Best Answer
Case 1
$a_n\leq \frac{1}{2^n}$ for all $n\in \mathbb N$
Then $na_n\leq \frac{n}{2^n}$ for all $n\in \mathbb N$ and $\sum\frac{n}{2^n}$ converges by ratio test,so taking $M_n=n$ for all $n\in \mathbb N$ we are done.
Case 2
$a_n>\frac{1}{2^n}$ for some $n\in \mathbb N$
If there exist only finitely many $a_n$ such that $a_n>\frac{1}{2^n}$ then we are done by taking $M_n=n$ for all $n\in \mathbb N$.
If there is a subsequence $\{a_{r_{n}}\}$ such that $a_{r_{n}}>\frac{1}{2^{r_n}}$,then take $M_n$ as follows:
$M_n=\begin{cases} n ,\text{if } n\neq r_k\\\frac{1}{a_n^{1/n}},\text{if } n=r_k\end{cases}$
For $n=r_k$, $M_na_n=a_n^{1-\frac{1}{r_n}}<a_n$ and for $n\neq r_k$,$M_na_n=na_n\leq \frac{n}{2^n}$ So, $\sum M_na_n$ is convergent.