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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!!!

Note by Sharky Kesa
3 years, 1 month ago

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

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Numberphile!!!! Kartik Sharma · 2 years, 10 months ago

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@Kartik Sharma No, Wikipedia. Sharky Kesa · 2 years, 10 months ago

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@Sharky Kesa oh!

I learnt it from that... Kartik Sharma · 2 years, 10 months ago

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Weird...I posted a discussion on these exact sums... Bogdan Simeonov · 3 years, 1 month ago

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@Bogdan Simeonov when, because I want to read it. Sharky Kesa · 3 years, 1 month ago

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@Sharky Kesa Here it is Interesting sums Bogdan Simeonov · 3 years, 1 month ago

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@Bogdan Simeonov Wow. I found this when I was surfing the internet. Sharky Kesa · 3 years, 1 month ago

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