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!!!
Comments
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Top NewestThis is a famous result that \(\zeta (-1) = -\dfrac{1}{12}\). (Saw this in Number Phile :P). – Surya Prakash · 1 year, 7 months ago
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Numberphile!!!! – Kartik Sharma · 3 years, 2 months ago
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No, Wikipedia. – Sharky Kesa · 3 years, 2 months ago
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oh!
I learnt it from that... – Kartik Sharma · 3 years, 2 months ago
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Weird...I posted a discussion on these exact sums... – Bogdan Simeonov · 3 years, 5 months ago
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when, because I want to read it. – Sharky Kesa · 3 years, 5 months ago
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Here it is Interesting sums – Bogdan Simeonov · 3 years, 5 months ago
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Wow. I found this when I was surfing the internet. – Sharky Kesa · 3 years, 5 months ago
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