Another Fibonacci series

Let $F_n$ denote the $n$ -th Fibonacci number. Evaluate the sum:

$$\mathcal{S} = \sum_{n=0}^\infty \sum_{k=0}^n \frac{F_{2k}F_{n-k}}{10^n}$$

Solution

Well this is honestly nothing once you know some things about the generating functions of Fibonacci. All we need are the following formulae:

$$\sum_{k=0}^{\infty}F_{2k}x^k = \frac{x}{1-3x+x^2} \tag{1}$$

and

$$ \sum_{k=0}^{\infty}F_{k} x^k = \frac{x}{1-x-x^2} \tag{2}$$

Using the Cauchy product and multiplying $(1)$ and $(2)$ we get that:

\begin{align*}
\left (\sum_{k=0}^{\infty} F_{2k} x^k   \right )\left ( \sum_{k=0}^{\infty} F_k x^k \right ) &= \sum_{n=0}^{\infty} \left ( \sum_{k=0}^{n} F_{2k} F_{k-n} \right )x^n \\
 &=\frac{x^2}{\left ( 1-3x+x^2 \right )\left ( 1-x-x^2 \right )}
\end{align*}

Thus for $x=\frac{1}{10}$ our desired sum evaluates to $\mathcal{S}=\frac{100}{6319}$.