Skip to main content
edited tags
Link
Martin Sleziak
  • 4.7k
  • 4
  • 35
  • 40
added 1 character in body
Source Link
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

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 if true.

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 $n\in\mathbb{Z}$.

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, I like to ask for alternative proofs if true.

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 $n\in\mathbb{Z}$.

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 $n\in\mathbb{Z}$.

Source Link
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

Fibonacci series captures Euler $e=2.718\dots$

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, I like to ask for alternative proofs if true.

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 $n\in\mathbb{Z}$.