Let $\{a_n\}$ be a sequence with $a_n\to l\ne0$ as $n\to\infty$.

Then, does the series $$\sum^\infty_{n=1}(a_n-a_{n+1})$$ necessarily converge?

I don’t know which of the following arguments is correct:

The series converges:

Because $$\sum^N_{n=1}(a_n-a_{n+1})=a_1-a_N$$ Taking limit $N\to\infty$ on both sides, the series converges to $a_1-l$.

The series diverges:

Set $a_n=1$ for all $n$.

Then the series is equivalent to $$1-1+1-1+1-1\cdots$$ which does not converge.

Furthermore, if the series does not always converge, under what conditions would the series converge?

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    $\begingroup$ Guess the sums are indexed by $n$, not by $k$. The series converges Yes, iff $a_n$ converges. The series diverges But you are not allowed to remove the parentheses, unless stricter conditions are met. Keeping the parentheses in place, the latter series is just $(1-1)+(1-1)+\ldots = 0+0+\ldots$ $\endgroup$ – dxiv Aug 14 '18 at 3:40
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    $\begingroup$ If $a_n=1$ for all $n$, then all of the terms in your series are $0$, not alternating $\pm 1$! $\endgroup$ – hmakholm left over Monica Aug 14 '18 at 3:41
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    $\begingroup$ Just a small comment to illustrate the fallacy of your second argument: "one can rewrite $1/2^n = 1/2^n + 1 - 1$, so $\sum_n 1/2^n = \sum_n (1/2^n + 1 - 1)$ must diverge, by the same argument." $\endgroup$ – Clement C. Aug 14 '18 at 5:47
  • $\begingroup$ Important thing to know that is applicable to all series: If $a_n$ does not tend to $0$ then $\sum a_n$ diverges. Doesn't matter what type of series it is. $\endgroup$ – David Reed Sep 15 '18 at 23:49

You must realize that the infinite series (which is rigorously denoted by a summation expression) is defined as the limit of the partial sums if it exists, and the partial sums are in turn defined in a certain precise way. One is not permitted to do any rearrangement, especially by 'intuition', unless one proves that doing so does not change the limit of the partial sums. This is why $\sum_{k=0}^\infty (-1)^k$ diverges while $\sum_{k=0}^\infty ((-1)^{2k}+(-1)^{2k+1})$ converges.

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  • $\begingroup$ I would just add here that if the series converges absolutely than it can always be rearranged in any manner without changing its value, and if it does not, that is if it conditionally converges, then it can actually be rearranged to sum to any number you want. $\endgroup$ – David Reed Sep 15 '18 at 23:46

You are assuming we don't know how the series is structured. But we do - instead of having $$1-1+1-1+\cdots$$ we instead have $$(1-1)+(1-1)+\cdots$$ so every term cancels and the series clearly converges! This is more evident from the sigma notation, since $a_n = a_{n+1} = 1$ so $\sum (a_n - a_{n+1}) = \sum 0 = 0$

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