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Manipulating Infinite Sums

!THE FOLLOWING MATH IS HIGHLY ILLOGICAL! Hey guys, I've been talking to my friends Josh Speckman and Nate Ji

So, you've probably seen the \(1+2+3+4+5 \cdots = - \dfrac{1}{12}\). If not, here: We will define three sums: \(S _1 = 1-1+1-1+1-1 \cdots\), \(S _2 = 1-2+3-4+5-6 \cdots\), and \(S _3 = 1+2+3+4+5 \cdots\).

Looking at the first sum, we see that \(1-S _1=1-(1-1+1-1+1 \cdots) = 1-1+1-1+1-1 \cdots\). Thus, \(S _1= \dfrac{1}{2}\).

Now we must solve the second sum. We get \(S _2 +S _2 = 2S _2 = 1 -2+3-4+5 \cdots + 1 - 2 + 3 - 4 + 5 \cdots = 1 + (- 2 + 1) + (3-2) + (-4+3) \cdots = 1-1+1-1+1-1 \cdots = \dfrac{1}{2}\). Thus \(2S _2 = \dfrac{1}{2} \Rightarrow S _2 = \dfrac{1}{4}\).

Now the last sum. We see that \(S _3 - S _2 = 1+2+3+4+5 \cdots - (1 - 2 + 3 - 4 + 5 - 6 \cdots) = (1-1) + (2+2) +(3-3) + (4+4) \cdots = 4 + 8 + 12 + 16 \cdots = 4S _3\). So \(S_3 - \dfrac{1}{4} = 4S _3 \Rightarrow 3S _3 = - \dfrac{1}{4} \Rightarrow S _3 = - \dfrac{1}{12}\).

Now, I was showing my friends this, and they decided to continue along the theme of not-quite valid mathematics. So, here it is:

\(1+2+3+4+5\cdots\) is the sum of the first \(\infty\) positive integers, so it is \(\dfrac{\infty(\infty + 1)}{2} = \dfrac{\infty^2 + \infty}{2}\). Since this is equal to \(-\dfrac{1}{12}\), we get \(\dfrac{\infty^2+\infty}{2} = - \dfrac{1}{12}\),

so \(\infty^2+\infty=-\dfrac{1}{6} \Rightarrow \infty^2+\infty+\dfrac{1}{6} = 0 \),

and \(\infty = \dfrac{-1 \pm \sqrt{\dfrac{1}{3}}}{2} \Rightarrow \infty =\boxed{ \dfrac{-3 \pm \sqrt{3}}{6}}\).

Yay infinite divergent series!

Note by Frodo Baggins
3 years ago

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I DON'T UNDERSTAND YOU. Tanisha Martheswaran · 3 years ago

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Well, infinity squared is still infinity and there are different types of infinity so you are practically stating that the divergent sum (zeta of -1) is equal to infinity AND minus one twelfth, and both things can be true. Bogdan Simeonov · 3 years ago

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