## Series 1/5+1/25+2/125+3/625+…

Compute
$S=\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{2}{125}+\dfrac{3}{625}+\dfrac{5}{3125}+\cdots$, where each numerator after the second is the sum of the two preceding numerators and each denominator is $5$ times the preceding one.
Source: NCTM Mathematics Teacher 2006

SOLUTION
$5S=1+\dfrac{1}{5}+\dfrac{2}{25}+\dfrac{3}{125}+\dfrac{5}{625}+\dfrac{8}{3125}+\cdots$
$S=\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{2}{125}+\dfrac{3}{625}+\dfrac{5}{3125}+\cdots$
Subtract $S$ from $5S$
$4S=1+\dfrac{1}{25}+\dfrac{1}{125}+\dfrac{2}{625}+\dfrac{3}{3125}+\cdots$
$=1+\dfrac{1}{5}S$
Solving for $S$
$\dfrac{19S}{5}=1$
$S=5/19$

Answer: $5/19$