Sequences $(\lambda_n)$ such that for every summable sequence $(a_n)$, $(\lambda_na_n)$ is also summable. An oral examination exercise :

Find all the sequences $(\lambda _n)_n$ of real numbers such that :
  $$\sum a_n \; \text{is convergent} \Longrightarrow \sum \lambda_n a_n \; \text{is convergent}$$

I think the only working sequences are the ones which are bounded and ultimately of constant sign.
I've not seen it on the site so if you have seen it please tell me how you would solve it. 
Thanks. 
 A: The sequences of bounded variation, we must have
$$\sum_{n = 0}^\infty \lvert \lambda_{n+1} - \lambda_n\rvert < +\infty.$$
For a convergent series $\sum a_n$, consider the sequence of tail sums
$$r_n = \sum_{k = n}^\infty a_k.$$
Then $(r_n)$ is a sequence converging to $0$, and with $a_n = r_n - r_{n+1}$ a summation by parts yields
\begin{align}
\sum_{n = 0}^N \lambda_n a_n &= \sum_{n = 0}^N \lambda_n (r_n - r_{n+1})\\
&= \sum_{n = 0}^N \lambda_n r_n - \sum_{n = 1}^{N+1} \lambda_{n-1} r_n\\
&= \lambda_0 r_0 - \lambda_N r_{N+1} + \sum_{n = 1}^N (\lambda_{n} - \lambda_{n-1}) r_n.
\end{align}
It is easy to see that the boundedness of $(\lambda_n)$ is necessary, so with that restriction we have $\lambda_N r_{N+1} \to 0$, and
$$\sum_{n = 0}^\infty \lambda_n a_n$$
is convergent if and only if
$$\sum_{n = 1}^\infty (\lambda_n - \lambda_{n-1})r_n\tag{1}$$
is convergent. Since $\sum \lambda_n a_n$ shall converge for all convergent $\sum a_n$, $(1)$ must converge for all $r \in c_0$, where $c_0$ is the Banach space of sequences converging to $0$, endowed with the supremum norm. The topological dual of $c_0$ is $\ell^1$, and by the Banach-Steinhaus theorem, a sequence $(\mu_n)$ such that
$$\lim_{N\to\infty} \sum_{n = 1}^N \mu_n\cdot r_n$$
exists for all $r\in c_0$ belongs to $\ell^1$.
Conversely, it is elementary to see that all sequences $(\lambda_n)$ of bounded variation have the desired property.
A: If $\lambda_n$ is not bounded consider the sequence
$$a_n=\begin{cases}{}\sum_{j=p}^q\frac1{j^2}&\text{ if }\lambda_n<p\le q\le\lambda_{n+1}\\0&\text{ otherwise}\end{cases}$$
