This problem has an interesting history from the early days of real analysis attached to it. I really like this story and the proof it contains of the problem in this question is also different from the other answers posted.
In 1827 a M. Oliver published a proof in Crelle's Journal (Journal für die reine und angewandte Mathematik) of the following theorem:
Theorem M. Oliver: A series of positive terms converges if and only if $\lim_{n\to\infty} na_n = 0$.
Not long after this paper had been published Niels Henrik Abel responded with a short note noting that the theorem "does not seem to hold true" as the series $\sum_{n=2}^\infty\frac{1}{n\log(n)}$ diverges but $\lim_{n\to\infty}n\cdot \frac{1}{n\log(n)} = 0$ contradicting the theorem above.
Abel then followed it up by another short note showing that there cannot exist any function $\phi(n)$ such that
$$\sum_{k=1}^\infty a_n ~~~\text{converges} \iff \lim_{n\to\infty}\phi(n)a_n = 0$$
The proof is remarkably simple and uses the lemma below (which coincides with the problem this question asks for). I will here present Abel's original proof. It's from memory so it's not word for word but the main idea behind the proof should be intact.
Lemma: if a series of positive terms $\sum_{k=1}^\infty a_k$ diverges then so does the series $\sum_{k=1}^\infty \frac{a_k}{a_1+a_2+\ldots + a_{k-1}}$.
Proof: Let $s_n = \sum_{k=1}^{n}a_k$, then
$$\log\left(\frac{s_k}{s_{k-1}}\right) = \log\left(1+\frac{a_k}{s_{k-1}}\right) < \frac{a_k}{s_{k-1}}$$
since $\log(1+x) < x$. Summing over $k=2,3,\ldots,n$ the left hand side telescopes to give
$$\log\left(\frac{s_n}{s_{1}}\right) < \sum_{k=1}^n \frac{a_k}{s_{k-1}}$$
Now if $s_n = \sum_{k=1}^n a_k$ diverges then so does the logarithm of $s_n$ and it follows that $\sum_{k=1}^n \frac{a_k}{s_{k-1}}$ also diverges.
Proof of Abel's theorem above: Assume there exist a function $\phi(n)$ with the properties that $\sum a_k$ converges if and only if $\lim_{n\to\infty}\phi(n)a_n = 0$. Then $\sum_{k=1}^\infty \frac{1}{\phi(k)}$ diverges since $\lim_{n\to\infty} \phi(n)\frac{1}{\phi(n)} = 1$. By the lemma above this implies that $$\sum_{k=1}^\infty \frac{\frac{1}{\phi(k)}}{\frac{1}{\phi(1)}+\frac{1}{\phi(2)} + \ldots + \frac{1}{\phi(k-1)}}$$
also diverges but
$$\lim_{n\to\infty} \phi(n)\frac{\frac{1}{\phi(n)}}{\frac{1}{\phi(1)}+\ldots + \frac{1}{\phi(n-1)}} = \lim_{n\to\infty}\frac{1}{\frac{1}{\phi(1)}+\ldots + \frac{1}{\phi(n-1)}} = 0$$
which would indicate that it is convergent. No such function $\phi(n)$ can therefore exist.