Show that if $a_n$ is descending and $ \sum a_n$ is convergent then $na_n\to 0$ Show that if $a_n$ is descending and $\sum a_n$ is convergent then $na_n \to 0$.
Does $\sum a_n$ have to be convergent?
I see that, but I can't prove it. I think that it need not to be convergent, but i also can't prove it.
 A: Given $\epsilon>0$,the sequence $S_n:=\sum_{i=1}^n u_i$ converges to $S:=\sum_{i=1}^\infty u_i$, 
there exists a natural number N such that $|S - S_n|<\epsilon$ implying $\sum_{i>N} u_i<\epsilon$. 
For $n>2N$ we have : 
$\frac{n}{2}u_n\leq(n-N)u_n\leq u_{N+1}+...+u_n\leq\sum_{i>N} u_i<\epsilon$
A: Use the Cauchy criterion, that is, that for any $\varepsilon >0$ we can find $M$ such that $m,n>M$ gives $$\tag 1 |a_{n}+\cdots+a_m|<\varepsilon$$
To show $na_n\to 0$, of course also using that $a_n\geqslant a_{n+1}$. 
Hint Show that $na_{2n}\to 0$ and $n a_{2n+1}\to 0$ separately by a good choice of $n$ and $m$ in $(1)$.
A: The method in the previous answer is probably the best, but here is a slightly different approach using the same ideas:
Given $\epsilon>0$, choose an N such that if $n\ge m\ge N$, then $\displaystyle\lvert\sum_{k=m}^{n}a_k\rvert=\sum_{k=m}^{n}a_k<\frac{\epsilon}{2}$.
$\textbf{Hint}$ Now use that If $n\ge N$, then
$a_N+a_{N+1}+\cdots+a_n<\frac{\epsilon}{2}$ where
$a_N+a_{N+1}+\cdots+a_n\ge a_n+a_n+\cdots+a_n$.
$---------------------------------------$
Here is another method:
Since $\displaystyle\sum_{n=1}^{\infty}a_n$ converges and $(a_n)$ decreases to 0,
$\displaystyle\sum_{n=1}^{\infty} 2^n a_{2^n}$ converges by the Cauchy Condensation Test.
Therefore $2^na_{2^n}\rightarrow0$, so $na_n\rightarrow0$ by the Squeeze Theorem.
(See $\lim_{n\to\infty}a_n$ and Cauchy condensation)
