Proof for bracketing a positive series

Question

I've been asked to prove that given a series consisting of non-negative terms. By bracketing these terms in a particular way, I get a new series that converges. Prove that the original series converges also and to the same sum.

I know that it is fairly obvious that if the bracketed series is convergent, and all the terms are positive, surely the original series must be convergent, but having trouble formalizing an actual proof for that.

Is it really as simple as since the bracketed series is absolutely convergent, which in a sense "removes the brackets". Therefore the original series is absolutely convergent and hence convergent? — dahaka5, Feb 18 '17 at 13:10
When you say "bracketing", do you mean rearranging or just grouping terms without changing order? — MPW, Feb 18 '17 at 13:19

score 2 · Accepted Answer · answered Feb 18 '17 at 13:03

This comes from the fact that the partial sum sequence is increasing ($U_{n+1}-U_n=u_{n+1}\ge0$), so it either converges or is not bounded. The bracketed sum gives you a extracted sequence $(U_{\phi(n)})_n$ from the sequence $(U_n)_n$, but for monotonous sequences, the boundedness of one is equivalent to the boundedness of the other (easy exercise).

Proof for bracketing a positive series

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