TITLE: Infinite geometric sum (asking for insight on an easier solution)
QUESTION [12 upvotes]: 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.

REPLY [4 votes]: $$S(x)=\sum_{n=1}^{\infty} x^n F_n=xF_1+x^2 F_2+ \sum_{n=3}^{\infty} x^n F_n=$$ $$=x+x^2+\sum_{n=3}^{\infty} x^n (2F_{n-1}+F_{n-2})=$$ $$=x+x^2 +2x\sum_{m=2}^{\infty}x^m F_m+x^2\sum_{m=1}^{\infty}xF_m=$$ $$=x+x^2 + 2x(S(x)-xF_1)+x^2S(x)=x+x^2+2x(S(x)-x)+x^2S(x).$$ $$\text {Therefore  } \quad S(x)(1-2x-x^2)=x+x^2-2x^2=x-x^2.$$  When $x=0.1$ this implies $S(0.1)\cdot (0.79)=0.09.$