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!!!

#Summation #NumberFallacy

Easy Math Editor

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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
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Comments

Weird...I posted a discussion on these exact sums...

Bogdan Simeonov - 7 years, 4 months ago

when, because I want to read it.

Sharky Kesa - 7 years, 4 months ago

Here it is Interesting sums

Bogdan Simeonov - 7 years, 3 months ago

@Bogdan Simeonov – Wow. I found this when I was surfing the internet.

Sharky Kesa - 7 years, 3 months ago

Numberphile!!!!

Kartik Sharma - 7 years ago

No, Wikipedia.

Sharky Kesa - 7 years ago

oh!

I learnt it from that...

Kartik Sharma - 7 years ago

This is a famous result that $\zeta (-1) = -\dfrac{1}{12}$ . (Saw this in Number Phile :P).

Surya Prakash - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$