# Two sums with Fibonacci numbers

1. Find closed form formula for sum: $\displaystyle\sum_{n=0}^{+\infty}\sum_{k=0}^{n} \frac{F_{2k}F_{n-k}}{10^n}$

2. Find closed form formula for sum: $\displaystyle\sum_{k=0}^{n}\frac{F_k}{2^k}$ and its limit with $n\to +\infty$.

First association with both problems: generating functions and convolution. But I have been thinking about solution for over a week and still can't manage. Can you help me?

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for the second part $\sum _{k=0}^{\infty }\dfrac {F^{k}} {p^{k}}=\dfrac {p} {p^2 - p-1}$ for $p\geq 0$ source en.wikipedia.org/wiki/Fibonacci_number#Power_series –  Hardy Aug 7 '12 at 10:41
See this for the second part. –  Ｊ. Ｍ. Aug 7 '12 at 11:14

For (2) you have $F_k = \dfrac{\varphi^k}{\sqrt 5}-\dfrac{\psi^k}{\sqrt 5}$ where $\varphi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$ so the problem becomes the difference between two geometric series.

For (1) I think you can turn this into something like $\displaystyle \sum_{n=0}^{\infty} \frac{F_{2n+1}-F_{n+2}}{2\times 10^n}$ and again make it into a sum of geometric series.

There are probably other ways.

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I don't know which identity to use to simplify $\sum_{k=0}^n F_{2k}F_{n-k}$ –  ray Aug 7 '12 at 15:01
@ray: You could try to prove $\sum_{k=0}^n F_{2k}F_{n-k} = \frac{F_{2n+1}-F_{n+2}}{2}$ by induction –  Henry Aug 7 '12 at 15:57
I know, but it isn't well known identity, I think. At least I couldn't find it on wiki. I'm wondering how it can be deduced. –  ray Aug 7 '12 at 16:05
@ray: work out the first few terms, look it up on OEIS and find A056014 which says "Convolution of Fibonacci numbers F(n) with F(2n)" –  Henry Aug 7 '12 at 18:17
thank you. Unfortunately OEIS will not help me during the exam or something like that. But I came up with slightly different solution and I think as good as yours. Indeed it is a convolution of F(2n) with F(n) so all we need is to find generating functions for both sequences which is easy. Then multiply them and the answer is its value for $z=1/10$. I like finding more than one solution. Thanks once again. –  ray Aug 7 '12 at 19:15