Let $(a_n)$ be any sequence and $(b_n)=n(a_n-a_{n+1})$.

Prove that if $\sum a_n$ and $\sum b_n$ converges then $\lim_{n\to \infty}na_n=0$ and $\sum a_n= \sum b_n$.

The second part, assuming the first part, I've shown. 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 $0$.

  • $\begingroup$ You mean absolute convergence, no? $\endgroup$ – Hasan Saad May 9 '15 at 14:59
  • $\begingroup$ I think the problem is correct. I don't think if you set $a_n$ as above then $\sum b_n$ converges. $\endgroup$ – nomadicmathematician May 9 '15 at 15:02
  • $\begingroup$ Sorry for overlooking the condition on $b_n$. You are right. $\endgroup$ – Alex Fok May 9 '15 at 15:03
  • $\begingroup$ Summation by parts $\endgroup$ – lvb May 9 '15 at 15:05

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 $$ (N+1)\:a_{N+1}=\sum_{n=1}^{N+1} a_{n}-\sum_{n=1}^{N} b_n, \quad N\geq1. $$

  • $\begingroup$ This shows that the limit $(na_n)$ exists, but how do I get that it is equal to zero? $\endgroup$ – nomadicmathematician May 9 '15 at 15:14
  • $\begingroup$ I showed that the limit must be zero. Please check if it's correct. $\endgroup$ – nomadicmathematician May 9 '15 at 15:23
  • $\begingroup$ Your proof is correct. But you dont need the $\liminf$ and $\limsup$ in there. $\endgroup$ – Tim B. May 9 '15 at 15:26
  • $\begingroup$ @takecare Well done! Thanks. $\endgroup$ – Olivier Oloa May 9 '15 at 15:27

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 $0$.

  • $\begingroup$ I showed that the limit is $0$, please check if my solution is correct. $\endgroup$ – nomadicmathematician May 9 '15 at 15:24

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