The Fibonacci recurrence $F_n=F_{n-1}+F_{n-2}$ allows values for all indices $n\in\mathbb{Z}$. There is an almost endless list of properties of these numbers in all sorts of ways. The below question might even be known. Yet, if true, I like to ask for alternative proofs.

Question. Does the following identity hold? We have $$\frac{\sum_{k=0}^{\infty}\frac{F_{n+k}}{k!}}{\sum_{k=0}^{\infty}\frac{F_{n-k}}{k!}}=e,$$ independent of $n\in\mathbb{Z}$.

  • $\begingroup$ Perhaps this is also true with F replaced by G, where G is a Fibonacci recurrence but G_0 and G_1 are two arbitrary numbers, excluding 0,0. Gerhard "Can't Build E From Nothing" Paseman, 2017.04.12. $\endgroup$ – Gerhard Paseman Apr 12 '17 at 21:04
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    $\begingroup$ Can't you just do the sums using the Binet formula? $\endgroup$ – Aaron Bergman Apr 12 '17 at 21:05
  • $\begingroup$ As I mentioned above, it would be nice to see alternative or novel proofs. $\endgroup$ – T. Amdeberhan Apr 12 '17 at 21:19
  • $\begingroup$ @GerhardPaseman I suspect your comment is true, and can be proved from the special cases $n=0,1$ of the OP's assertion by taking linear combinations. $\endgroup$ – Greg Martin Apr 13 '17 at 17:07

It follows from the identity

$$F_{n+k} = \sum_{j=0}^k {k \choose j} F_{n-j}$$

which is obtained by applying the standard recurrence $k$ times to the left side, each time splitting up each term into two terms.

Indeed, this gives

$$\sum_{k=0}^{\infty} \frac{F_{n+k}}{k!} = \sum_{k=0}^{\infty} \frac{ \sum_{j=0}^k {k \choose j} F_{n-j}}{k!}= \sum_{k=0}^{\infty} \sum_{j=0}^{k} \frac{1}{j! (k-j)!} F_{n-j} = \sum_{j=0}^{\infty} \sum_{k-j=0}^{\infty} \frac{1}{(k-j)!} \frac{F_{n-j}}{j!} = \sum_{j=0}^{\infty} e \frac{F_{n-j}}{j!}$$

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    $\begingroup$ OoOooO that is slick! $\endgroup$ – user78249 Apr 12 '17 at 22:21
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    $\begingroup$ This is cool, too. $\endgroup$ – T. Amdeberhan Apr 12 '17 at 22:38
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    $\begingroup$ It is not an answer, It's a Golden answer! $\endgroup$ – Amin235 Apr 18 '17 at 21:44

Mathematica tells me it's a consequence of the two series (distinguished by $\pm$): $$\sum_{k=0}^\infty\frac{F_{n\pm k}}{k!}=\frac{e^{\sqrt{5}} \phi^n-(1-\phi)^n}{\sqrt{5}\, \exp(\phi^{\mp 1})},$$ with $\phi$ the golden ratio (the positive solution of $1+1/\phi=\phi$). So the ratio of the $\pm$ series equals $e^{\phi-1/\phi}=e$, and the entire $n$-dependence drops out of the ratio.

The identity also holds for $n\in\mathbb{R}$, with the usual generalisation $F_z=(\phi^z-\phi^{-z}\cos\pi z)/\sqrt{5}$ of the Fibonacci numbers to real argument $z$.

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    $\begingroup$ This is cool, indeed. $\endgroup$ – T. Amdeberhan Apr 12 '17 at 22:37
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    $\begingroup$ As I suspected (and Greg Martin commented), this extends to general nontrivial Fibonacci recurrences with $G_0=a$ and $G_1=b$. Writing the RHS above as $f(n)$, the corresponding value using $G$ is $af(-1) + bf(0)$ for the positive version, and similarly for the negative. Again cancellation occurs leaving a power of $e$. Gerhard "Or Use General Linear Nonsense" Paseman, 2017.04.13. $\endgroup$ – Gerhard Paseman Apr 13 '17 at 18:04

More generally, $$\sum_{k=0}^\infty F_{n+k} \frac{x^k}{k!} = e^x\sum_{k=0}^\infty F_{n-k}\frac{x^k}{k!},\tag{1}$$ which is equivalent to Will Sawin's identity.

Similarly, $$e^x\sum_{k=0}^\infty F_{n+k}\frac{x^k}{k!}= \sum_{k=0}^\infty F_{n+2k}\frac{x^k}{k!},\tag{2}$$ and more generally, from the formula $$(-1)^q F_{p-q}+F_{\!q}\,\phi^p = F_{\!p}\,\phi^q,$$ where $\phi = (1+\sqrt5)/2$, and the analogous formula with $\phi$ replaced by $(1-\sqrt5)/2$, we get for any integers $p$, $q$, and $n$, $$ e^{(-1)^q F_{p-q}\ x}\sum_{k=0}^\infty F_{q}^k F_{n+pk}\frac{x^k}{k!} = \sum_{k=0}^\infty F_p^k F_{n+qk} \frac{x^k}{k!}. $$ The cases $p=1, q=-1$ and $p=1,q=2$ give $(1)$ and $(2)$.

Related identities (also involving Lucas numbers) can be found in L. Carlitz and H. H. Ferns, Some Fibonacci and Lucas Identities, Fibonacci Quarterly 8 (1970), 61–73.


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