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The following identity appears in Martin Gardner's paper, "Dr. Matrix on Little Known Fibonacci Curiosities:

$$e = \frac{1 + 1 + \frac{2}{2!} + \frac{3}{3!} + \frac{5}{4!} + \frac{8}{5!} + \frac{13}{6!} + \frac{21}{7!} + \frac{34}{8!} + \frac{55}{9!} + \cdots}{1 + 0 + \frac{1}{2!} + \frac{1}{3!} + \frac{2}{4!} + \frac{3}{5!} + \frac{5}{6!} + \frac{8}{7!} + \frac{13}{8!} + \frac{21}{9!} + \cdots} $$

How can we prove this?

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This may be helpful .$F_n=\dfrac{\phi^n+\psi^n}{\sqrt{5}}$ where $x^2-x-1=(x-\psi)(x-\phi)$ –  Ishan Banerjee Apr 13 '13 at 15:30
    
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If we interpret this as $$\frac{\sum\limits_{k=0}^\infty \tfrac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty \tfrac{F_{k-1}}{k!}}$$ then I don't think this claim is correct; the result I get is $$\frac{1+(1+\phi)\exp(\sqrt 5)}{1+\phi+\exp(\sqrt 5)}\approx 2.1291534094361927319$$ –  J. M. Apr 13 '13 at 16:40

1 Answer 1

up vote 10 down vote accepted

The correct form of this expression seems to be: $$e=\frac{1+1+\frac{2}{2!}+\frac{3}{3!}+\frac{5}{4!}+\frac{8}{5!}+\frac{13}{6!}+\ldots}{1-0+\frac{1}{2!}-\frac{1}{3!}+\frac{2}{4!}-\frac{3}{5!}+\frac{5}{6!}-\ldots} = \frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}}$$

There are two useful observations that can be used to prove this statement:

  • $\sum\limits_{k=0}^\infty \frac{x^k}{k!}=e^x$ (the definition of $e^x$)
  • $F_n=\frac{\varphi^n-\psi^n}{{\varphi-\psi}}$, where $\varphi=\frac{1+\sqrt{5}}{2}$ and $\psi=\frac{1-\sqrt{5}}{2}$ (the Euler-Binet formula)

Now we are all set for the proof! $$\begin{eqnarray} \sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{\varphi^{k+1}-\psi^{k+1}}{k!} \\ & = & \frac{1}{\varphi-\psi}\left(\varphi\sum\limits_{k=0}^\infty \frac{\varphi^k}{k!}-\psi\sum\limits_{k=0}^\infty \frac{\psi^k}{k!}\right) \\ & = & \frac{1}{\varphi-\psi}\left(\varphi e^\varphi - \psi e^\psi\right) \\ \end{eqnarray}$$

$$\begin{eqnarray} \sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!} & = & \frac{1}{\varphi-\psi}\sum\limits_{k=0}^\infty \frac{(-1)^k\varphi^{k-1}-(-1)^k\psi^{k-1}}{k!} \\ & = & \frac{1}{\varphi-\psi}\left( \frac{1}{\varphi}\sum\limits_{k=0}^\infty \frac{(-\varphi)^k}{k!} - \frac{1}{\psi}\sum\limits_{k=0}^\infty \frac{(-\psi)^k}{k!}\right) \\ & = & \frac{1}{\varphi-\psi}\left(\frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi}\right) \\ \end{eqnarray}$$

Dividing these two yields: $$\frac{\sum\limits_{k=0}^\infty \frac{F_{k+1}}{k!}}{\sum\limits_{k=0}^\infty (-1)^k\frac{F_{k-1}}{k!}} = \frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} }$$

Since $\varphi + \psi = 1$ and $\varphi\psi = -1$, we this can be simplified as $$\frac{\varphi e^\varphi - \psi e^\psi}{ \frac{1}{\varphi} e^{-\varphi} - \frac{1}{\psi} e^{-\psi} } = \frac{\varphi e^\varphi - \psi e^\psi}{ e^{-1}\left(\varphi e^{\varphi} - \psi e^{\psi}\right)} = e$$

Q.E.D.

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Nice proof! (+1) –  Carl Najafi May 24 '13 at 8:47
    
A wonderful answer! Thanks for the correction. –  01000100 May 24 '13 at 9:52

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