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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.

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    For $\lim_n a_nb_n = \lim_n a_n\cdot \lim_n b_n$ we need the limits of $a_n$ and $b_n$ to exist. – clark May 26 '18 at 23:26
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    "Since $u_n$ converges by assumption, we have $\lim\limits_{n \to \infty} u_n = 0$" Wait, isn't that what you want to show? – imranfat May 26 '18 at 23:26
  • "$\lim\limits_{n \to \infty} nu_n = \lim\limits_{n \to \infty} n \cdot \lim\limits_{n \to \infty} u_n = \lim\limits_{n \to \infty} n \cdot 0 = 0$" The first and last equality, or the first and second, depending on how you read $\lim\limits_{n \to \infty} n \cdot 0 $, are very, very wrong. – Arnaud Mortier May 26 '18 at 23:30
  • @imranfat: I thought this followed from the fact that $\sum u_n$ converged; we wanted to show that $\lim n \cdot u_n$ is $0$ as well. –  May 26 '18 at 23:31
  • @Arnaud Mortier: Sorry, I had a feeling that wouldn't be clear. I was trying to use the convergence of $u_n$ to conclude that $\lim\limits_{n \to \infty} u_n = 0$ and replace this with $0$, but I think clark's comment renders this incorrect, as I forgot that we can only use this by assuming the limit of $n$ exists. –  May 26 '18 at 23:33
  • @Matt.P My comment is an extension of Clark's comment - which I hadn't seen. What he says is not the only big mistake in that line. – Arnaud Mortier May 26 '18 at 23:34
  • My apologies; I don't think I see it. Could it be that $\lim\limits_{n \to \infty} n = \infty$, and $0 \cdot \infty$ is an indeterminate form? –  May 26 '18 at 23:45

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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 $0$.

robjohn
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  • what if the limit does not exist? (Well, ok, the OP assumes the limit does exist. But that assumption seems unnecessary.) – Mirko May 26 '18 at 23:33
  • Thank you for this answer. I definitely see where I went wrong in my incorrect application of the limit-product rule. Would you mind explaining, though, the logic behind the reasoning that $n u_n \geq \frac{L}{2}$? I'm having some difficulty understanding where this came from. –  May 26 '18 at 23:35
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    @Mirko yes, it is. Pick $u_n=\frac1n$ if $n$ is a perfect square, and $0$ otherwise. – Arnaud Mortier May 26 '18 at 23:36
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    @Matt.P think $(L-\varepsilon,L+\varepsilon)$ where $\varepsilon=\frac L2$. – Mirko May 26 '18 at 23:37
  • @ Mirko: Thanks for this, though I still don't think I'm following. It seems that this probably follows from the definition of the limit, right? –  May 27 '18 at 02:05
  • @robjon you are right..i did not read the post carefully...you are correct then..+1 from me. – Marios Gretsas May 27 '18 at 02:34
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    @MariosGretsas: this answer addresses the question you were thinking about. – robjohn May 27 '18 at 02:52
  • @Matt.P: The definition of $\lim\limits_{n\to\infty}nu_n=L$ : $$\forall\epsilon\gt0,\exists N:\forall n\ge N,|nu_n-L|\le\epsilon$$ Set $\epsilon=L/2$, then using the $N$ guaranteed by the existence of the limit, we get for $n\ge N$, that $|nu_n-L|\le L/2$, which, if $L\gt0$, implies by the triangle inequality, that $nu_n\ge L/2$. – robjohn May 27 '18 at 03:03
  • Thank you for this; one last question on this, if you wouldn't mind. (My apologies for not seeing this.) I understand that we can make this difference as small as we want. (I've always written the definition with $< \epsilon$, but I assume, for our purposes with this proof, it makes no difference). But, I'm struggling to see how we get from $\lvert nu_n - L \rvert \leq \frac{L}{2}$ to $nu_n \geq \frac{L}{2}$. The triangle inequality gives us an expression $\leq$ $\lvert nu_n - L \rvert$ and lets us remove the absolute-value bars, but I can't seem to show that it's less than $\frac{L}{2}$. –  May 27 '18 at 03:40
  • Disregard the above comment -- I found it. Thank you very much! –  May 27 '18 at 03:46
  • @Matt.P: just in case anyone else has the same question, for $L\gt0$, $$\begin{align}nu_n &=L+(nu_n-L)\ &\ge L-|nu_n-L|\ &\ge L/2\end{align}$$ – robjohn May 27 '18 at 07:10