I've been working quite a bit on this proof and reading some hints people have posted on here, and I think I have pieced together a possible proof. I'd appreciate if someone could look this over and verify it. I'm worried there could be some leaps of logic. The theorem is out of Schaum's Advanced Calculus book.
Theorem: If $\sum u_n$ converges, where $u_n \geq 0$ for $n > N$, and if $\lim\limits_{n \to \infty} n u_n$ exists, prove that $\lim\limits_{n \to \infty} n u_n = 0$.
Proof. Assume, for a contradiction, that $\lim\limits_{n \to \infty} nu_n \neq 0$. Since it exists by assumption, it must equal some finite number, say, $L$. Since $u_n$ converges by assumption, we have $\lim\limits_{n \to \infty} u_n = 0$. Using the limit-product rule, we have \begin{align*} \lim\limits_{n \to \infty} nu_n = \lim\limits_{n \to \infty} n \cdot \lim\limits_{n \to \infty} u_n = \lim\limits_{n \to \infty} n \cdot 0 = 0 \end{align*} which is a contradiction. Thus, $\lim\limits_{n \to \infty} nu_n$ must equal $0$.
It seems I could be making excessie lepas here, and I don't use the assumption that $u_n \geq 0$, though I do use convergence. I would really appreciate any helpful comments people may have on this.
Comment on the Solution
Note that $$ \lim_{n\to\infty}nu_n=\lim_{n\to\infty}n\lim_{n\to\infty}u_n $$ is only true if the limits on the right exist. However, $\lim\limits_{n\to\infty}n$ does not exist.
Note that, by the reasoning in the question, if $u_n=\frac1n$, then we would have $$ 1=\lim_{n\to\infty}\overbrace{nu_n\vphantom{\lim_{n\to\infty}}}^1=\lim_{n\to\infty}n\overbrace{\lim_{n\to\infty}u_n}^0=0 $$
Answer
If $\lim\limits_{n\to\infty}nu_n=L\gt0$, then there is some $N$ so that for $n\ge N$, we have $nu_n\ge\frac L2$. Then $$ \sum_{n=N}^\infty u_n\ge\sum_{n=N}^\infty\frac L{2n} $$ which diverges.
If $\lim\limits_{n\to\infty}nu_n=L\lt0$, the series diverges similarly, even if we don't have $u_n\ge0$.
Thus, if the limit exists, it must be $0$.