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Let $$S \triangleq \sum_{k=0}^\infty\,(-1)^k\quad.$$ On one hand, the sequence of partial sums alternates between $0$ and $1$, therefore, does not get arbitrarily close to any value, and $S$ can't converge. On the other hand, it's easy to see that $$S = (-1)^0 + \sum_{k=1}^\infty\,(-1)^k = 1 + \sum_{k=0}^\infty\,-(-1)^k=1-S\quad\therefore\quad S = \frac12\quad,$$ so that $S$ is finite. One of these rationales has to be wrong, right?

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up vote 1 down vote accepted

You get the same result by applying the formula for a geometric series:

\begin{array}{ccc} S &=& a + ar + ar^2 + \dots \\ rS &=& ar + ar^2 + ar^3 + \cdots \end{array}

Hence $S - rS = a$ and so $S = \frac{a}{1-r}$. If you put $a=1$ and $r=-1$ then you get $S=\frac{1}{2}$. The problem is that $S$ is only well-defined if $|r|<1$. If $|r| \ge 1$ then $S$ is not well-defined. You can't do arithmetic with something that is not well-defined. Your final answer does not make sense because you have been manipulating objects that do not make sense.

You can do all sorts of crazy things when working with infinity. For example: $\infty + \infty = \infty$, so dividing through by $\infty$ gives $1+1 =1$.

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Your first argument is correct. The second contains a built-in assumption that $S$ exists, which is not the case.

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Indeed, $S$ neither converges nor diverges. – Luke May 13 '13 at 19:50
@Luke: Actually, $S$ does diverge: to say that a series diverges is simply to say that it does not converge. I suspect that you mean that its partial sums do not increase (or decrease) without bound. – Brian M. Scott May 13 '13 at 20:13
Yes. I thought a real divergent sequence had to go to $\pm\infty$. – Luke May 14 '13 at 0:43

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