Suppose $S = 1 - 1 + 1 - 1 + 1 - 1 + 1\ldots$. Then $1 - S = S$ so $S = \frac{1}{2}$ Now suppose $T = 1 - 2 + 3 - 4\ldots$. We can also write $T = 0 + 1 - 2 + 3 - 4\ldots$. Adding the two we get $2T = (1+0)+(-2+1)+(3-2)+(-4+3)\ldots$ or $2T = 1 - 1 + 1 - 1\ldots$, so $T = \frac{1}{4}$. Now let $P=1+2+3+4+5\ldots$. It can be easily seen that $P-T = 4P$ or $P=-\frac{T}{3} =-\frac{1}{12}$ so we have $P = -\frac{1}{12}$ How it is possible that $1 + 2 + 3 + 4\ldots=-\frac{1}{12}$?
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2Here's some info http://math.stackexchange.com/q/39802/115823 – turkeyhundt Feb 13 '15 at 04:39
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You cannot actually define $S$ because it doesn't even converge. – Vim Feb 13 '15 at 04:39
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@Vim One can define anything; the difficult part is making a consistent definition. :) There are certain ways of assigning values to "classically" divergent series, such as Cesàro summation. – apnorton Feb 13 '15 at 04:45
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@anorton However I don't see the point in doing this o(╯$\nabla$╰)o – Vim Feb 13 '15 at 10:27
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This comes down to a fundamental misconception of an infinite series. The series $\sum_{i=0}^\infty (-1)^i$ diverges, so the notation $1-S = S$ doesn't make any sense. Similarly for the other claims. It turns out infinity is weird, and so adding together infinitely many numbers is weird too.
Further, from our common understanding of the convergence of an infinite series, $\sum_{i=i}^\infty i$ diverges, as in the sequence of partial sums $\{ \sum_{i=1}^k i\}_{k \in \mathbb{N}}$ has infinite limit.
walkar
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