For convergent series of positive terms $\exists \psi$ with $\lim_{s \to \infty} \psi(s)=+\infty$ but $\sum_1^\infty \psi(n)a_n < \infty$ . The problem is to show that for a series of positive terms, $a_n \geq 0$ with 
$\sum_1^\infty a_n < \infty$, there exists a continuous function $\psi: \textbf{R}^+ \to \textbf{R}^+$ with $\lim_{s \to \infty} \psi(s)=+\infty$ but $\sum_1^\infty \psi(n)a_n < \infty$. I know that for specific sequences a function can usually be found (e.g., for $a_n=\frac{1}{n^3}, \psi(s)=s$ would work)  but don't know how to give a constructive or non-constructive proof for the general case.
 A: David's comment gives a link to an excellent answer by Sami Ben Romdhane. Here is a slightly more general result than the one given by Sami.
For any $n\in\mathbb N$, set $$R_n:=\sum_{k=n}^\infty a_n\, .$$
Note that the sequence $(R_n)$ is decreasing with $\lim_{n\to\infty} R_n=0$.
Let us prove the following
Fact. For any decreasing function $f:(0,R_1]\to \mathbb R^+$ such that $\int_0^{R_1} f(t)\, dt<\infty$, the series $\sum f(R_n)a_n$ is convergent.
Assuming this has been proved, take any function $f$ as above such that  $\lim_{t\to 0^+} f(t)=\infty$; for example, $$f(t)=\frac{1}{t^\alpha}\,,\; \alpha\in (0,1)\, .$$ Then $f(R_n)$ increases to $\infty$ as $n\to\infty$. So the function $\psi:\mathbb R^+\to\mathbb R^+$ such that $\psi(n)=f(R_n)$ for all $n\in\mathbb N$ and $\psi$ is affine on each interval $[n-1,n]$, has the required properties.
To prove the fact, note that for any $n\in\mathbb N$, we have 
$$ f(R_n)\, {a_n}=f(R_n)\, ({R_n-R_{n+1}})\leq \int_{R_{n+1}}^{R_n}f(t)\, dt\, , $$
because $f(t)\geq f(R_n)$ on the interval $[R_{n+1},R_n]$. It follows that 
$$\sum_{n=1}^\infty f(R_n)\, {a_n}\leq \sum_{n=1}^\infty \int_{R_{n+1}}^{R_n}f(t)\, dt=\int_0^{R_1}f(t)\, dt<\infty\, .$$
