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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
2 years, 10 months ago

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I DON'T UNDERSTAND YOU. · 2 years, 10 months ago