<s>[INST] Infinite geometric sum (asking for insight on an easier solution) Let the sequence $F$ be defined as: $F_1=F_2=1$ and $F_n=2F_{n-1}+F_{n-2}$, for $n>2$. Evaluate $\sum_{n=1}^{\infty}\frac{F_n}{10^n}$.
The obvious solution involves solving for the explicit formula for $F$ (using the standard linear recurrence technique): $F_n=(\frac{\sqrt{2}-1}{2})(1+\sqrt{2})^n-(\frac{\sqrt{2}+1}{2})(1-\sqrt{2})^n$. Then we can just split the sum into two infinite geometric series. The computation is annoying, but nonetheless straightforward.
I was wondering if there is any easier solution than solving for an explicit formula for $F$. Maybe there is one that only needs the recursive definition? Any comments are appreciated.
 [/INST] \begin{align*}
\sum_{n=1}^\infty\frac{F_n}{10^n}&=\frac{1}{10}+\frac{1}{100}+\sum_{n=3}^\infty\frac{F_n}{10^n}\\
&=0.11+\sum_{n=3}^\infty\frac{2F_{n-1}+F_{n-2}}{10^n}\\
&=0.11+\frac{2}{10}\left(\sum_{n=1}^\infty\frac{F_n}{10^n}-\frac{1}{10}\right)+\frac{1}{100}\sum_{n=1}^\infty\frac{F_n}{10^n}\\
0.79\sum_{n=1}^\infty\frac{F_n}{10^n}&=0.09\\
\sum_{n=1}^\infty\frac{F_n}{10^n}&=\frac{9}{79}
\end{align*}
</s>