Summation (Again!!!)

To read my previous note on this click here

What does $1 + 2 + 3 + 4 + 5 + 6 + ...$ equal?

We need to work with other identities first.

Let $S_1 = 1 - 1 + 1 - 1 + ...$;

and $S_2 = 1 - 2 + 3 - 4 + ...$.

First we solve $S_1$.

$S_1 = 1 - 1 + 1 - ...$

$1 - S_1 = 1 - (1 - 1 + 1 - ...)$

$1 - S_1 = 1 - 1 + 1 - ...$

$1 - S_1 = S_1$

$S_1 = \frac {1}{2}$

Now we solve $S_2$.

$S_2 = 1 - 2 + 3 - 4 + ...$

$2S_2 = (1 - 2 + 3 - 4 + ...) + (1 - 2 + 3 - 4 + ...)$

By shifting the addend one value to the right you get

$2S_2 = 1 - 1 + 1 - 1 + ...$

By using our previous value, we get

$2S_2 =\frac {1}{2}$

$S_2 = \frac {1}{4}$

Finally, we solve $S$

$S = 1 + 2 + 3 + 4 + ...$

$S - S_2 = (1 + 2 + 3 + 4 + ...) - (1 - 2 + 3 - 4 + ...)$

$S - S_2 = 4 + 8 + 12 + ...$

$S - S_2 = 4 (1 + 2 + 3 + ...)$

$S - \frac {1}{4} = 4S$

$S = -\frac {1}{12}$

Ta da!!!