## Insane Sums !

A) $$\sum_{n=0}^{\infty }(-1)^{n} = 1-1+1-1+1-... =$$ $$\frac{1}{2}$$

B) $$\sum_{n=1}^{\infty }(-1)^{n+1}n = 1-2+3-4+5-... =$$ $$\frac{1}{4}$$

C) $$\sum_{n=1}^{\infty }(2)^{n} = 1+2+4+8+16+... = -1$$

D) $$\sum_{n=1}^{\infty }n = 1+2+3+4+5+... = \frac{-1}{12}$$

E) $$\sum_{n=1}^{\infty }n^2 = 1^2+2^2+3^2+4^2+5^2+... = 0$$

F) $$\sum_{n=1}^{\infty }n^3 = 1^3+2^3+3^3+4^3+5^3+... =$$ $$\frac{1}{120}$$

## Proofs:

A)
Let $$S$$ be the sum $$\sum_{n=0}^{\infty }(-1)^{n}$$  such that  $$S =1-1+1-1+1-...$$ Then, by subtracting the sum $$S$$ from number $$1$$ will yield:
$$1-S = 1-(1-1+1-1+....)$$ $$= 1-1+1-1+...= S$$
Thus $$1-S = S$$  $$\Rightarrow$$ $$S=\frac{1}{2}$$
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B)
Let $$S$$ be the  Cesàro sum $$\sum_{n=1}^{\infty }(-1)^{n+1}n$$ such that $$S= 1-2+3-4+5-6+...$$ Then, by adding the sum $$S$$ to itself with a slight shift will yield the following results:
$$\left.\begin{matrix} \begin{matrix} &1 &-2 &+3 &-4 &+5 &-6 &+... \\ +& & & & & & & \\ & &1 &-2 &+3 &-4 &+5 &-... \end{matrix} \end{matrix}\right\}$$  $$\Rightarrow$$ $$2S=1-1+1-1+...$$ So, From A) we knew that this sum is equal to $$\frac{1}{2}$$.

Hence, $$2S=\frac{1}{2}$$ $$\Rightarrow$$  $$S=\frac{1}{4}$$
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C)
Let $$S$$ be the sum of the binary base system $$\sum_{n=1}^{\infty }(2)^{n}$$ such that $$S= 1+2+4+8+16+...$$ Then, by manipulating this term, the sum $$S$$ can be expressed by:
$$S= 1+2+4+8+16+...$$
$$= 1+2(1+2+4+8+16+...)$$
$$= 1+2S$$
Thus, $$S=1+2S$$ $$\Rightarrow$$ $$S=-1$$

If you are trying to imagine the sense behind this sum, we recommend you watching the 3Blue1Brown video below:
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D)
Let $$S$$ be the sum of the natural numbers $$\sum_{n=1}^{\infty }n$$ such that  $$S= 1+2+3+4+5+...$$ Then, multiplying the Sum $$S$$ by $$4$$ and subtract it from $$S$$ itself considering the shifting strategy yields:
$$\left.\begin{matrix} \begin{matrix} &1 & +2 & +3 & +4 & +5 & +6 & +... \\ - & & & & & & & \\ & &4 & &+8 & & +12 & +... \end{matrix} \end{matrix}\right\}$$ $$\Rightarrow$$ $$-3S=1-2+3-4+5-6+...$$

Thus, $$-3S=\frac{1}{4}$$ $$\Rightarrow$$ $$S=\frac{-1}{12}$$

This Sum often called the Golden Nugget, and you may watch the video below on Numberphile's to know why ?
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F)
Let $$S$$ be the sum of the cubic natural numbers $$\sum_{n=1}^{\infty }n^3$$ such that $$S=1^3+2^3+3^3+4^3+5^3+...$$ Then,
Thus,   $$\Rightarrow$$ $$S=\frac{1}{120}$$
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## Discussions

Two points of view can be considered from these sums:
1. Some people may ask, what sense behind this infinite sums? And in fact, scientists have made effort to make the emerged sums worthy. In physics, they utilize the Sum E in and in string theory. For instance, in  Physicsbuzz.Physicscentral.com, they stated that "The Numberphiles point out that the relation $$1+2+3+4+ . . .= \frac{-1}{12}$$ (though not true) is useful for string theory because strings are one-dimensional." Also, F comes up in the calculation of the three dimensional Casimir Effect problem. Thus, these insane sums are related in somehow to real world, which make us more convinced about their existence and validity.

2. In another way, we may notice that B, D, E, & F,  depends on A in their proofs. However, A is still skeptic since if we look at it in this manner;   Let $$S=1-1+1-1+....$$ then $$2S=S+S = S$$ $$\Rightarrow$$ $$2S=S$$ $$\Rightarrow$$ $$2=1$$ which is not true in our mathematical logic and conception.
Evantually, the main questions remained:
• Why these calculations related to some constants, but other calculations leave out to Infinity?
• Is there another mathematical logic we shall follow when dealing with infinity.
• What is the right approach for overcoming this sort of sums.