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!
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?
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.
