Ramanujan Sum

Definition

Assigning values to diverging series is called Ramanujan summation, and it is represented through the symbol $\Re$.

Theorem

[1] The Grandi Series ¹: $$ 1-1+1-1+ \cdots = {{ 1 } \over { 2 }} \qquad ( \operatorname{Re} ) $$
[2] $$ 1-2+3-4+ \cdots = {{ 1 } \over { 4 }} \qquad ( \operatorname{Re} ) $$
[2]’ $$ 1+2+3+4+ \cdots = - {{ 1 } \over { 12 }} \qquad ( \operatorname{Re} ) $$

Explanation

One might wonder what it means to assign values to something that does not converge, let alone diverges. For now, it might be adequate to think of it as formally calculating the result as if it were converging. Surprisingly, such methods are applied in fields like physics, even if we don’t fully understand or feel the necessity to specifically search and explain them. Therefore, only intuitive (not rigorous) proofs are introduced here. The proofs themselves are interesting to look at, and I hope readers will view them with interest and not take them too seriously.

In particular, for theorem [2]’, a more rigorous proof was introduced in a separate post through the analytical continuation of the Dirichlet series $\displaystyle \sum_{n \in \mathbb{N}} {{ a_{n} } \over { n^{s} }}$, which is the Riemann zeta function. If a reader has not understood the rigorous proof, it would be difficult to answer questions about the Ramanujan summation. Interestingly, the results are the same whether the proof is intuitive or rigorous, but intuitive proofs have many weaknesses, and it is too complicated to explain rigorous proofs in just a few words.

Proof

[1]²

If we set it as $$ S := 1-1+1-1+ \cdots \qquad ( \operatorname{Re} ) $$

then $$ 1 - S = 1 - \left( 1-1+1-1+ \cdots \right) = 1-1+1-1+ \cdots = S \qquad ( \operatorname{Re} ) $$

Moving $S$ from the left side to the right side gives us $$ 1 = 2S \qquad ( \operatorname{Re} ) $$

Therefore $$ 1+1+1+1+ \cdots = S = {{ 1 } \over { 2 }} \qquad ( \operatorname{Re} ) $$

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[2]³

If we set it as $$ S := 1 - 2 + 3 - 4 + 5 - 6 + \cdots \qquad ( \operatorname{Re} ) $$

then $$ \begin{align*} S =& 1 & - & 2 & + & 3 & - & 4 & + & 5 & - & 6 & + & \cdots \qquad ( \operatorname{Re} ) \\ 2S =& & & 2 & - & 4 & + & 6 & - & 8 & + &10 & - & \cdots \qquad ( \operatorname{Re} ) \\ S =& & & & & 1 & - & 2 & + & 3 & - & 4 & + & \cdots \qquad ( \operatorname{Re} ) \end{align*} $$

Therefore $$ 4S := 1 + 0 + 0 + \cdots \qquad ( \operatorname{Re} ) $$

Therefore $$ 1 - 2 + 3 - 4 + \cdots = {{ 1 } \over { 4 }} $$

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[2]'

If we set it as $$ S := 1 + 2+3+4+ \cdots \qquad ( \operatorname{Re} ) $$

then $$ \begin{align*} S =& 1 & + & 2 & + & 3 & + & 4 & + \cdots \qquad ( \operatorname{Re} ) \\ 4S =& & & 4 & + & & + & 8 & + \cdots \qquad ( \operatorname{Re} ) \\ S - 4S =& 1 & - & 2 & + & 3 & - & 4 & + \cdots \qquad ( \operatorname{Re} ) \end{align*} $$

According to theorem [2] and because of $\displaystyle S = {{ 1 } \over { 4 }}$,

$$ 1+2+3+4+ \cdots = - {{ 1 } \over { 12 }} \qquad ( \operatorname{Re} ) $$

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