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I couldn't think of a good title for this, given the nature of the proof; if anyone has a better idea, feel free to edit (I think that's possible).

I'm trying to prove this following theorem:

If $\sum u_n$ converges, where $u_n \geq 0$ f or $n > N$, and if $\lim\limits_{n \to \infty} n u_n$ exists, prove that $\lim\limits_{n \to \infty} nu_n = 0$.

I'm having some difficulty putting together this proof. I think we may want to start with the fact that since $\sum u_n$ converges, we have that $\lim\limits_{n \to \infty} u_n = 0$. With some algebraic manipulation, we can convert this to a limit of the form $\frac{0}{0}$ to which we can apply L'Hospital's Rule, but this doesn't seem to help much. I would assume that the assumption that each term of the series, except for possibly some finite number, are positive might come in here, but this doesn't tell us anything about the derivative, though the series should, technically, be decreasing after some point and bounded below by $0$ in order to converge, so we should be able to write a negative derivative, $u_n'$, for an infinite number of terms. I haven't been able to work out a solution with L'Hospital's rule, though, and the limit product rule, breaking apart this product, isn't conclusive, as I'm fairly certain $\infty \cdot 0$ is an indeterminate form.

The other option might be to proceed by contradiction, though I'm a bit lost as to how to proceed from here. I'd greatly appreciate any helpful insights people may have on this.

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Let $\lim_{n \to \infty} n u_n = L$ and note $L \ge 0$.

Suppose $L > 0$. Then there exists $N'$ such that $u_n \ge \frac{L}{2n}$ for all $n \ge N'$. What does this imply about $\sum_{n = N'}^\infty u_n$?

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  • $\begingroup$ If I understand this correctly, it seems you're proceeding by contradiction. (If I'm wrong on this, the rest of my comment isn't relevant.) But, why then would we take $L \geq 0$ rather than $\neq 0$? Is the argument that we can rule out a negative $L$ because $u_n \geq 0$, and now we only need to rule out $L > 0$? From here, I'm having some difficulty following where the $\frac{1}{2n}$ came from. $\endgroup$ Commented May 26, 2018 at 19:22
  • $\begingroup$ Yes, by contradiction (or alternatively, by contrapositive). Yes, to justify $L \ge 0$ I am using the fact that $u_n \ge 0$ for large $n$, so we only need to rule out $L > 0$ as you said. Lastly, since $n u_n \to L$, we have $n u_n \ge L/2$ for all sufficiently large $n$. $\endgroup$ Commented May 26, 2018 at 19:32
  • $\begingroup$ My apologies, but I've been thinking this over quite a bit and I can't seem to grasp where the $\frac{L}{2}$ comes from. Would you mind explaining this argument further? $\endgroup$ Commented May 27, 2018 at 2:04
  • $\begingroup$ @Matt.P The choice of $L/2$ is arbitrary; anything strictly smaller than $L$ suffices. The intuition is that as $n$ gets large, $n u_n$ gets closer and closer to $L$. Although we can't say that $n u_n \ge L$ for all large $n$, we can still say $n u_n \ge L(1-\epsilon)$ for any $\epsilon > 0$. $\endgroup$ Commented May 27, 2018 at 3:01
  • $\begingroup$ So, if I understand correctly, we're tending toward L, and can make the difference between $u_n$ and $L$ arbitrarily small for a sufficiently large $N$, so we can make it small enough by choosing some $N$ s.t. $n u_n < \frac{L}{2}$. I think this makes sense. (I'm not sure if I understand the $L(1 - \epsilon)$ argument you've written; wouldn't this only hold if $\epsilon$ is between $0$ and $1$?) From here, is the argument that $\sum \frac{L}{2n}$ diverges, as it's a multiple of the harmonic series, so by the comparison test, $u_n$ diverges, a contradiction? $\endgroup$ Commented May 27, 2018 at 3:30

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