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

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About mvtrinh

Retired high school math teacher.
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