I was presented with the following problem;

Show that if $\sum b_n$ is a rearrangement of a series $\sum a_n$ , and $a_n$ diverges to $\infty$, then $\sum b_n = \infty$.

How would one solve this? It seems intuitively true, but how could I show it?

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    $\begingroup$ The series $1+1-1+1+1-1+\dots$ diverges to infinity, but $1-1+1-1+\dots$ simply oscillates. $\endgroup$ – lulu Mar 9 '16 at 11:51
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    $\begingroup$ The series 1+1-1+1-1 does not diverge to infinity. It diverges, but not to infinity. $\endgroup$ – Dr. John A Zoidberg Mar 9 '16 at 11:52
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    $\begingroup$ OP, How is a rearrangement of an infinite series defined? Because the first series @lulu gives does diverge to infinity. But to me it is not trivial $1+1−1+1+1−1+\ldots$ is a rearrangement of $1−1+1−1+\ldots$, as you are messing with the numbers of $-1$ and $+1$ terms... $\endgroup$ – Eric S. Mar 9 '16 at 11:58
  • $\begingroup$ @Aex. the partial sum, $S_n$ of that series is bounded below by $\lfloor \frac n3\rfloor$. To be clear, the series is defined by $a_n=-1$ if $3|n$, $a_n=1$ otherwise. $\endgroup$ – lulu Mar 9 '16 at 11:58
  • $\begingroup$ @lulu: are you sure you obtain the first from the second by simply rearranging the terms? $\endgroup$ – Alex Mar 9 '16 at 12:09

I suppose you mean that $a_n\to +\infty$. Under this assumption, we can proceed as follows.

Let $M>0$ be an arbitrary number. Then there is some $N>0$ such that $a_n>1$ for any $n>N$. Since $\{b_n\}$ is a rearrangement of $\{a_n\}$, then there is some $N'>0$ such that $a_1,\ldots,a_N$ are contained in $b_1,\ldots,b_{N'}$.

Set $B=\sum_{k=1}^{N'}b_k$. Then for any $T>|B|+M+N'$, we have $$\sum_{k=1}^{T}b_k=\sum_{k=1}^{N'}b_k+\sum_{k=N'+1}^{T}b_k\ge B+(T-N')>M.$$

Therefore, $\sum b_n\to+\infty$.

If we only know $\sum a_n=+\infty$, we can say nothing about $\sum b_n$. See Riemann rearrangement theorem.


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