Two sums with Fibonacci numbers 
  
*
  
*Find closed form formula for sum: $\displaystyle\sum_{n=0}^{+\infty}\sum_{k=0}^{n} \frac{F_{2k}F_{n-k}}{10^n}$
  
*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?
 A: The  first  one can  be  solved using  the  fact  that the  generating
function of the Fibonacci numbers is
$$\frac{z}{1-z-z^2}.$$
Introduce the function
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n \frac{F_{2k} F_{n-k}}{10^n}$$
so that we are interested in $f(1).$
Re-write $f(z)$ as follows:
$$f(z) = \sum_{k\ge 0} F_{2k} 
\sum_{n\ge k} \frac{z^n}{10^n} F_{n-k}
= \sum_{k\ge 0} F_{2k} \frac{z^k}{10^k} 
\sum_{n\ge 0} \frac{z^n}{10^n} F_n.$$
Now we have
$$ \sum_{k\ge 0} F_{2k} z^{2k} =
\frac{1}{2} \frac{z}{1-z-z^2}
- \frac{1}{2} \frac{z}{1+z-z^2}$$
and therefore $f(1)$ is
$$\left(\frac{1}{2} \frac{1/\sqrt{10}}{1-1/\sqrt{10}-1/10}
- \frac{1}{2} \frac{1/\sqrt{10}}{1+1/\sqrt{10}-1/10}\right)
\times 
\frac{1/10}{1-1/10-1/100}$$
which simplifies to
$$\frac{1}{2\sqrt{10}}
\frac{2/\sqrt{10}}{81/100-1/10} \times \frac{10}{89}
= \frac{1}{10} \times \frac{1}{71/100} \times \frac{10}{89} = 
\frac{100}{89\times 71}.$$
A: 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.
