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Let $a_k:=(-1)^k$ where $k\in\mathbb{N}$. $\mathbb{N}$ is the set of all non-negative integer.

And we define the partial sum $S_n:=\sum \limits_{k=0}^{n}a_k$. Notice that the sequence $\{S_k\}$ diverges which also implies that the infinite series $\sum \limits_{k=0}^{\infty}a_k$ cannot be defined.

If I consider the infinite sum $1-1+1-1+1-1+\cdots$, then this statement is equivalent that I just defined the infinite series $\sum \limits_{k=0}^{\infty}a_k$, which is contradiction.

But, suppose that the sum $1-1+1-1+1-1+\cdots$ exists and let the value of the sum be $S$. Then, we can easily observe that $S=1-S$, therefore $S=1/2$. The supposition of this proposition already proved as false, but if I ignore the definition of infinite series, it holds.

Which one is right? Or are these propositions depends on what we define?

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1 Answers1

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Formally speaking, the series is divergent.

The sum of an infinite series is defined to be the limit of the sequence of its partial sums, if it exists. We see that the sequence of partial sums of the series, known as Grandi's series is $1, 0, 1, 0, …$, which does not approach any number (although has two accumulation points at $0, 1$). Therefore, we may conclude that Grandi's series (http://en.wikipedia.org/wiki/Grandi's_series) is divergent.

  • Despite the upvote and the hurried acceptance, the pedagogical usefulness of an answer stating without any qualm that $0=1=\frac12$ is not obvious to me. Everything before the last paragraph should be heavily redacted. – Did Mar 23 '14 at 17:02
  • @Did I am sorry, but I do not understand what you mean. I have tried my best to explain that the series diverges and that the "sum" of Grandi's series can be $0,1,$ or $0.5$. –  Mar 23 '14 at 17:05
  • I thought I was crystal clear but let me expand: you "compute" three times $1-1+1-1+\ldots$, to declare that this equals $0$, then $1$, then $\frac12$. Ergo, $0=1=\frac12$. Since the conclusion is absurd (do we agree on that?), something MUST be amiss in the "reasoning" leading to it (or ZF is contradictory). In other words, to play with these undefined sums might yield a kind of pleasure, but, to present such glib manipulations for mathematics is not a service one gives to anybody asking for explanations. – Did Mar 23 '14 at 17:13
  • @Did I will remove that part. –  Mar 23 '14 at 17:14
  • And then we run into the paradox that the answer that will stay is about 20% of the one that got accepted... But this remark rather concerns the asker/accepter, I guess. – Did Mar 23 '14 at 17:17