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If you expand $2^x$ using a finite difference series you end up with the formula

$$ 1 + x + \frac{1}{2!}x(x-1) + \frac{1}{3!}x(x-1)(x-2) ... = \sum_{n=0}^{\infty} \frac{(x)_n}{n!} $$

Now these series diverge for negative arguments, but they give some interesting results, i.e. they suggest

$$1 - 1 + 1 - 1 \ ... = \frac{1}{2}$$ $$1 - 2 + 3 - 4 \ ... = \frac{1}{4}$$ $$1 - 3 + 6 - 10 \ ... = \frac{1}{8} $$

and more generally... $$\begin{pmatrix} k \\ k \end{pmatrix} - \begin{pmatrix} k+1 \\ k \end{pmatrix} + \begin{pmatrix} k+2 \\ k \end{pmatrix} ... = 2^{-n}$$

Now what's funny... is that these divergent equalities, can actually be arrived at in a totally different way, which is by differentiating the function $ \frac{1}{1-x}$ $k$ times, dividing by k! and evaluating it at $x=-1$.

These two very distinct methods of infinite series seem to agree and so that has me wondering, is there a natural divergent summation method that encapsulates all this behavior? I want to say something like

$$2^x \equiv \sum_{n=0}^{\infty} \frac{(x)_n}{n!} \mod \text{summation method L }$$

$$ \frac{1}{1-x} \equiv \sum_{n=0}^{\infty} x^n \mod \text{summation method L} $$

this sort of abstract framework would be very useful and interesting to explore. But even something like Cessaro summation seems unsatisfactory, and it is not clear if Holder summation is sufficient for me. (I tried to calculate it with it and didn't end up getting the answer I expected).

Sidharth Ghoshal
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    $\lim_{z \to 1^-} \sum_{k=0}^\infty \frac{\prod_{l=0}^{k-1} (x-l)}{k!} z^k=\lim_{z \to 1^-} (1+z)^x= 2^x$. For $x = -n$ the result will be obviously the same as $\frac{1}{n!}\left. \frac{d^n}{dz^n}\frac{1}{1-z}\right|_{z=-1}$ – reuns Oct 16 '18 at 01:20
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    @reuns I think this deserves to be an answer. –  Oct 16 '18 at 01:56

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