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This question already has an answer here:

This came up in a friend's exam and it must be one of those ${\epsilon},N(\epsilon)$ arguments I could do in a snapshot in my twenties but now I can't figure out how the proof should go: For a positive non-increasing sequence $\{a_k, {k \in \mathbb{N}}\}$ if $$\sum_{k=1}^{\infty}a_k$$ converges then $b_n=na_n{\to}0$.

Thanks in advance!

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marked as duplicate by Did, Andrés E. Caicedo, Eric Stucky, MathOverview, Pedro Tamaroff Apr 5 '14 at 22:58

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If the sum is finite, then Cauchy-Criterion must be satisfied;

$\forall\epsilon>0, \exists\ n_0,n>n_0:\sum\limits_n^{2n}a_i<\epsilon/2$

Since $a_n\to0\Longrightarrow na_{2n}<\epsilon/2$

Hence $2(na_{2n})\to 0$

Now $a_{2n+1}\le a_{2n}$ and $(2n+1)a_{2n+1}\le (2n+1)a_{2n}=(\frac{2n+1}{2n})\underbrace{2\cdot n a_{2n}}_{\to 0}\to 0 $

Hence we've shown that $na_n\to 0$ if $n$ is odd or even.

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by the comparison test, if

$a_n \sim \frac{1}{n}$, then the series would diverge ($\sum \frac{1}{n}$ diverges)

Hence $a_n = o(\frac{1}{n}) \Rightarrow \lim_{n \to \infty} n\cdot a_n = \frac{a_n}{1/n} = 0$ by definition of $o(1/n)$

Edit

Suppose $$\lim n\cdot a_n = l$$ then

$l \neq 0$ is impossible, cause otherwise $a_n \sim \frac{l}{n}$ and diverges.

$l = \infty$ is impossible, because $a_n < \frac{1}{n}$ (otherwise $\sum a_n \ge \sum \frac{1}{n}$ that diverges) and so $na_n < 1$.

So if the limit exists it has to be $\lim n\cdot a_n = 0$.

To be fair I don't know how to prove that the limit exists; any suggestion?

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  • $\begingroup$ Not really. The negation of $a_n\sim 1/n$ is not $a_n=o(1/n)$, so how are you reaching that conclusion? $\endgroup$ – Andrés E. Caicedo Apr 5 '14 at 21:45
  • $\begingroup$ @AndresCaicedo take a look I edited :-) $\endgroup$ – Ant Apr 6 '14 at 11:03
  • $\begingroup$ Sure, that's better. Thanks. $\endgroup$ – Andrés E. Caicedo Apr 6 '14 at 15:03

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