The problem statement

Let $(a_n)_{n\geq o}$, $(Z_n)_{n\geq 0}$ sequences of complex numbers such that $(a_nZ_n)_{n\geq 0}$ converges. Show that

$\sum_{n=0}^{\infty} (a_n-a_{n+1})Z_n$ converges if and only if $\sum_{n=1}^{\infty} a_n(Z_n-Z_{n-1})$ converges.

I have no idea where to even start the problem, I am looking for any hints/suggestions to guide me.


Hint: $$ (a_{n}-a_{n+1})Z_{n}-a_{n+1}(Z_{n+1}-Z_{n})=a_{n}Z_{n}-a_{n+1}Z_{n+1}. $$

  • $\begingroup$ Let me see if this is correct: as the sequence $(a_nZ_n)_{n\geq 0}$ converges, then it is a Cauchy sequence, which means, $a_nZ_n-a_{n+1}Z_{n+1} \to 0$ when $n \to \infty$. But then $(a_{n}-a_{n+1})Z_{n}-a_{n+1}(Z_{n+1}-Z_{n}) \to 0$, this implies that the two sequences $(a_{n}-a_{n+1})Z_{n}$ and $a_{n+1}(Z_{n+1}-Z_{n})$ have the same limit. Now, I don't know how to show (if it is true) that if this happens, then the two series have the same limit. $\endgroup$ – user100106 May 3 '14 at 3:38
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    $\begingroup$ As a further hint, now sum both sides of the equation I gave you using $\sum_{n=0}^{N}$. $\endgroup$ – DisintegratingByParts May 3 '14 at 4:36
  • $\begingroup$ Oh, ok: First I call $\lim_{n \to \infty} a_nZ_n=a$, then $\sum_{n=0}^N a_nZ_n - a_{n+1}Z_{n+1}=a_0Z_0-a_{N+1}Z_{N+1}$, when $N \to \infty$, $a_0Z_0-a_{N+1}Z_{N+1} \to a_0Z_0-a$, but this implies $\sum (a_{n}-a_{n+1})Z_{n}=\sum a_{n+1}(Z_{n+1}-Z_{n})+(a_0Z_0 -a)$, from here one can deduce that one series converges if and only if the other does. $\endgroup$ – user100106 May 3 '14 at 5:02

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