Show that there is a sequence $(m_{j})_{j=0}^{\infty}$ s.t. $m_{j}\to\infty$ as $j\to\infty$ and $\sum_{j=0}^{\infty}m_{j}a_{j}$ converges. Suppose that $(a_{j})_{j=0}^{\infty}$ is a sequence of non-negative real numbers for which $\sum_{j=0}^{\infty}a_{j}$ converges. Show that there is a sequence $(m_{j})_{j=0}^{\infty}$ of positive real numbers such that $m_{j}\to\infty$ as $j\to\infty$ and $\sum_{j=0}^{\infty}m_{j}a_{j}$ converges.
MY ATTEMPT
I tried to consider $m_{j} = j$. Then we may apply the ratio test to the series $\sum_{j=0}^{\infty}m_{j}a_{j}$ in order to check for convergence:
\begin{align*}
\lim_{n\to\infty}\frac{(n+1)a_{n+1}}{na_{n}} = \lim_{n\to\infty}\left(\frac{n+1}{n}\right)\frac{a_{n+1}}{a_{n}}
\end{align*}
I also know that $a_{n}\to 0$ and $s_{n} = a_{1} + a_{2} + \ldots + a_{n} \leq M$, but then I got stuck.
Could someone help me with this?
 A: Let $b_j = a_j + 2^{-j}$ and $T_j = \sum_{k=j}^{\infty} b_k$. Also, define $m_j$ by
$$ m_j = \frac{\sqrt{T_j} - \sqrt{T_{j+1}}}{b_j} $$
Since $\sum_{j=0}^{\infty} b_j$ converges, $T_j$ converges to $0$ as $j\to\infty$. So
$$ \sum_{j=0}^{n} m_j a_j \leq \sum_{j=0}^{n} m_j b_j = \sum_{j=0}^{n} \bigl( \sqrt{T_j} - \sqrt{T_{j+1}} \bigr) \leq \sqrt{T_0} $$
shows that $\sum_{j=0}^{n} m_j a_j$ is bounded and hence converges. On the other hand,
$$ m_j = \frac{\sqrt{T_j} - \sqrt{T_{j+1}}}{b_j} = \frac{1}{b_j} \int_{T_j - b_j}^{T_j} \frac{\mathrm{d}x}{2\sqrt{x}} \geq \frac{1}{2\sqrt{T_j}} \xrightarrow[j\to\infty]{} \infty. $$
Therefore all desired conditions are satisfied.
A: Let $L=\sum_{n\ge 0}a_n$. For each $k\ge 0$ there is an $n_k\ge 0$ such that $\sum_{i\ge n_k}a_i\le\frac{L}{4^k}$, and we may further assume that $n_0=0$ and $\langle n_k:k\ge 0\rangle$ is strictly increasing. For $i\ge 0$ let $m_i=2^k$ iff $n_k\le i<n_{k+1}$; clearly $\lim_\limits{i\to\infty}m_i=\infty$. Then
$$\begin{align*}
\sum_{i\ge 0}m_ia_i&=\sum_{k\ge 0}\sum_{i=n_k}^{n_{k+1}-1}m_ia_i\\
&=\sum_{k\ge 0}2^k\sum_{i=n_k}^{n_{k+1}-1}a_i\\
&\le\sum_{k\ge 0}2^k\sum_{i\ge n_k}a_i\\
&\le\sum_{k\ge 0}2^k\left(\frac{L}{4^k}\right)\\
&=L\sum_{k\ge 0}\frac1{2^k}=2L\;.
\end{align*}$$
