To verify if this is a valid move?

Question

I am playing around with some numbers and it looks like this: $$\begin{align}e^{-3ix}&=\cos3x-i\sin3x=(\cos x-i\sin x)^3\\4\cos^3x-3\cos x-i(3\sin x-4\sin^3x)&=4(\cos^3x+i\sin^3x)-3e^{ix}\end{align}$$ Expanding $e^{-3ix}=(\cos x-i\sin x)^3=\cos^3x+i\sin^3ix-3i\cos ^2x\sin x-3\sin^2x\cos x\Rightarrow e^{-3ix}+3i\cos^2x\sin x+3\sin^2x\cos x=\cos^3x+i\sin^3x$

By substituting into the equation above, we get $$\begin{align}e^{-3ix}&=4(e^{-3ix}+3i\cos^2x\sin x+3\sin^2x\cos x)-3e^{ix}\\-e^{-3ix}&=4i\cos^2x\sin x+4\sin^2x\cos x-e^{ix}\\&=4\cos x\sin x(i\cos x+\sin x)-e^{ix}\\&=2\sin2x(\frac{-\cos x+i\sin x}{i})-e^{ix}\\&=2\sin2x\cdot(\frac{-e^{-ix}}{i})-e^{ix}\\&=2i\sin2x\cdot e^{-ix}-e^{ix}\end{align}$$ If we assume $e^{ix}=n$, then this becomes $n^{-3}=n-2n^{-1}i\sin2(i\ln n)$.
And since $-i\sin ix=\sinh x$ and we will take the sum from $n=1$ to infinity, we can make some changes to the equation above:
$$\begin{align}n^{-3}&=n-2n^{-1}\sinh (2\ln n)\\\sum_{n=1}^{\infty}n^{-3}&=\sum_{n=1}^{\infty}\frac {2\sinh(2\ln n)}{n}-\frac 1{12}\end{align}$$

My question is: if the series on the RHS has a special value although it diverges, then is it valid to find a real value in a complex way?

Answer 1

Your derivation seems ok, apart from a missed sign somewhere.

The equation $$n^{-3}=n+2n^{-1}\sinh (2\ln n)$$ should be $$n^{-3}=n+2n^{-1}\sinh (2\ln n)$$

at least according to Wolfram alpha.

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That said, your conclusion in the end is false. The two summand series,

$$\sum_{n=1}^\infty n\text{ and }\sum_{n=1}^\infty 2n^{-1}\sinh (2\ln n)$$

are both divergent series, so they do not have a sum, and you cannot conclude anything beyond that.

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This seems to stem from the common misconception that $$\sum_{n=1}^\infty n = -\frac1{12}$$ which is not true. The sum above does not exist, only its Ramanujan sum exists, and that one is indeed $\frac{-1}{12}$. But sums and Ramanujan sums are not the same thing, and cannot be used interchangeably.