×

# Summation (Again!!!)

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!!!

Note by Sharky Kesa
2 years, 8 months ago

Sort by:

This is a famous result that $$\zeta (-1) = -\dfrac{1}{12}$$. (Saw this in Number Phile :P). · 10 months, 3 weeks ago

Numberphile!!!! · 2 years, 5 months ago

No, Wikipedia. · 2 years, 5 months ago

oh!

I learnt it from that... · 2 years, 5 months ago

Weird...I posted a discussion on these exact sums... · 2 years, 8 months ago

when, because I want to read it. · 2 years, 8 months ago

Here it is Interesting sums · 2 years, 8 months ago