I am confused a bit while I am recalling the infinite series. Thomas' Calculus says: Sum of two divergent series can be convergent by giving an example: $\sum1 + \sum-1 = \sum0=0$. We also know that $\sum(-1)^n$ is divergent. However, can not we think the series $\sum(-1)^n=-1+1-1+1\cdots$ equals to $\sum1 + \sum-1$? What distinguishes these series exactly?

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    $\begingroup$ In $\sum 1 + \sum -1$, you can say that number of $1$'s and $-1$'s will be the same which is not in the case of $\sum(-1)^n$ $\endgroup$ – Isha Tarte Feb 27 '17 at 10:38
  • $\begingroup$ It seems that you are trying to re arrange the terms of $\sum (-1)^n$? $\endgroup$ – Juniven Feb 27 '17 at 10:44
  • $\begingroup$ @IshaTarte Yes, it seems that their numbers not equal and this leads my question actually. For the latter one, can not we correspond the $-1$'s and $1$'s one-to-one which means their numbers are equal? Why? $\endgroup$ – faith Feb 27 '17 at 10:57
  • $\begingroup$ @ΘΣΦGenSan Yes this is exactly what I try to. $\endgroup$ – faith Feb 27 '17 at 10:57
  • $\begingroup$ For your information, rearranging such terms is not allowed. See the answer bellow for a better explanation. $\endgroup$ – Juniven Feb 27 '17 at 11:08

Suppose that we have two sequences $\{a_n:n\ge1\}$ and $\{b_n:n\ge1\}$ and we want to find the limit of the sum of these two sequences. Then $$ \lim_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\lim_{n\to\infty}b_n $$ provided that both of the sequences $\{a_n:n\ge1\}$ and $\{b_n:n\ge1\}$ converge. If this is not the case, the equality might not hold.

Let us recall that a series is the limit of the sequence of the partial sums.

What you are actually doing in your example is the rearrangement of the terms of the series, which does not necessarily give you the same limit unless the series is absolutely convergent. By rearranging a conditionally convergent series you could actually get anything. This is the famous Riemann series theorem.

I hope this helps.


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