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With the geometric series:$$\sum_{n=0}^\infty r^n = \frac{1}{1-r} $$

It's normal to derive other formulas of series taking the derivative or integral of both sides of the formula. I wanted to see if taking the half-derivate of both sides could generate a new formula, but an error occurred and I don't know where.

Using the formulas from Half order derivative of $ {1 \over 1-x }$ I get:

$$\sum_{n=0}^\infty \frac{\Gamma(n+1)r^{n-\frac{1}{2}}}{\Gamma(n+\frac{1}{2})}=\frac{\sqrt{\pi}}{2(1-r)^{\frac{3}{2}}} $$

Moving the $r^{-\frac{1}{2}}$ to the other side:

$$\sum_{n=0}^\infty \frac{\Gamma(n+1)r^{n}}{\Gamma(n+\frac{1}{2})}=\frac{\sqrt{r\pi}}{2(1-r)^{\frac{3}{2}}} $$

Evaluating letting $r=\frac{1}{2}$ the sum = 2.0146... but the formula on the right = $\sqrt{\pi}$ which are clearly not the same. Am I making an algebraic mistake, is there something invalid about this intuition?

Tom Himler
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  • Do you have a specific reason to believe that the half-order derivative commutes with an infinite sum? – Umberto P. Feb 22 '18 at 17:14
  • @UmbertoP. nope, not at all. I would love a reason why it wouldn't work at all. I thought it should since it does for normal derivatives, but that intuition may not make sense in this situation. – Tom Himler Feb 22 '18 at 17:15
  • In fact, the fractional differential operator is neither commutative nor additive in general. – Mark Viola Feb 22 '18 at 17:24
  • @markviola That would explain the issue thanks. – Tom Himler Feb 22 '18 at 17:29

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