Given a convergent series there is divergent sequence such that $\sum a_n M_n$ is still convergent 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.
 A: 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.
A: The following theorem is proven here:

$\bf{Theorem: }$ Let $\sum_{n=1}^{\infty} c_n$ be any convergent
series with positive terms. Then, there exists a convergent series
$\sum_{n=1}^{\infty} C_n$ with much bigger terms in the sense that
$\lim_{n\rightarrow\infty} C_n/c_n = \infty.$

Given your sequence $\ \{a_n\},\ $ let $\ \{c_n\}=\{a_n\},\ $ then find $\ C_n\ $ as in the theorem, then let $\ M_n = C_n/c_n.$
A: Choose $A_n$ so $$\sum_{k>A_n}a_k<2^{-k}$$ and $A_{n+1}>A_n$. Define $$M_k=n\quad(A_n\le k<A_{n+1}).$$Then $$\sum M_ka_k=\sum_n\sum_{A_n\le k<A_{n+1}}M_ka_k\le\sum_n n2^{-n}<\infty.$$
A: Let $n_k$ be the first $n$ such that $\sum_{n\geq n_k}a_n<2^{-k}$, and let $M_n$ be the largest $k$ such that $n\geq n_k$. Since the original series is convergent, $M_n$ tends to $\infty$. On the other hand, $M_n$ grows slow enough that $\sum_{n=1}^{\infty}M_na_n$ converges, since each time $M_n$ increases by $1$, an exponentially decreasing quantity is added to the sum.
A: Sketch of the explicit construction:
Let $r_n=\sum_{j=n}^{\infty}a_j$. Clearly, $r_n$ is postive and converges to zero when $n\to\infty$. Now try to prove that $M_n=\frac{1}{\sqrt{r_n}+\sqrt{r_{n+1}}}$ works.
