Let $(a_n)$ be any sequence and $(b_n)=n(a_n-a_{n+1})$. If $\sum a_n$ and $\sum b_n$ converges then $\lim_{n\to \infty}na_n=0$ Let $(a_n)$ be any sequence and $(b_n)$ be a sequence such that $b_n := n(a_n-a_{n+1})$.
Prove that if $\sum a_n$ and $\sum b_n$ converges then $\lim_{n\to \infty}na_n=0$ and $\sum a_n= \sum b_n$.
I've shown the second part assuming the first part. I'm having trouble showing $\lim_{n\to \infty}na_n = 0$. I know that $\lim a_n=0$ and $\lim n(a_n-a_{n+1})=0$. How can I use these facts to show the first one?
I would greatly appreciate any help.
From the answers below I got that $\lim na_n$ exists. So I tried to show that the limit is $0$ by assuming that it is not.
First, assume that the limit $l \gt 0$. Then $\liminf na_n=l$. Hence for $l/2 \gt 0$, there is some $N$ such that for $n \ge N$, we have $na_n \gt l/2$. This implies that for $n \ge N$, $a_n \gt l/(2n)$. But this is a contradiction since we assumed that $\sum a_n$ converges.
Finally, assume that $l \lt 0$. Then $\limsup na_n=l$. So for $-l/2$, there is some $N$ such that if $n\ge N$ then $na_n\lt l/2$. This implies that $-a_n\gt -l/(2n)\gt 0$. Thus, again by comparison, we get that $-\sum a_n$ diverges, which is a contradiction.
Hence the limit must be $0$.
 A: Hint. You may write, for $N\geq1$,
$$
\begin{align}
\sum_{n=1}^{N} b_n&=\sum_{n=1}^{N} n(a_n-a_{n+1})\\\\
&=\sum_{n=1}^{N} n\:a_n-\sum_{n=1}^{N} n\:a_{n+1}\\\\
&=\sum_{n=1}^{N} n\:a_n-\sum_{n=1}^{N} (n+1)\:a_{n+1}+\sum_{n=1}^{N} a_{n+1}\\\\
&=-(N+1)\:a_{N+1}+\sum_{n=1}^{N+1} a_{n}
\end{align}
$$ or equivalently
$$
(N+1)\:a_{N+1}=\sum_{n=1}^{N+1} a_{n}-\sum_{n=1}^{N} b_n, \quad N\geq1.
$$
A: Let $A_n$ and $B_n$ be the $n$-th partial sum of $\sum_na_n$ and $\sum_nb_n$, respectively. Also dnote by $A$ and $B$ the sum of $\sum_na_n$ and $\sum_nb_n$, respectively. Then for every $n\ge 2$ we have:
\begin{eqnarray}
B_{n-1}&=&\sum_{k=1}^{n-1}k(a_k-a_{k+1})=\sum_{k=1}^{n-1}ka_k-\sum_{k=1}^{n-1}(k+1)a_{k+1}+\sum_{k=1}^{n-1}a_{k+1}\\
&=&\sum_{k=1}^{n-1}ka_k-\sum_{k=2}^nka_k+\sum_{k=2}^na_k=-na_n+\sum_{k=1}^na_k=A_n-na_n.
\end{eqnarray}
It follows that
$$
na_n=A_n-B_{n-1},
$$
and hence
$$
\lim_{n\to\infty}na_n=\lim_{n\to\infty}A_n-\lim_{n\to\infty}B_{n-1}=A-B,
$$
i.e. the sequence $(na_n)$ coverges, and its limit is $A-B$, not $0$.
