Sum includes powers of 5

Algebra Level pending

Compute :

$\frac{1}{5}+ \frac {1}{5^2} + \frac{2}{5^3}+ \frac{3}{5^4}+ \frac{5}{5^5}+ ... = \frac{A}{B}$

Each numerator is the sum of the two preceding numerators, and each denominator is $5$ times the preceding one.

What is $A+B =?$ (where $A$ and $B$ are coprime positive integers).

35 24 40 25 30

1 solution

Hana Wehbi
Mar 8, 2017

Let $\ S= \frac{1}{5}+\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{5}{5^5}+...$

then $\ 5\times S = 1+\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{5}{5^4}+...$

Subtract the series $\ S$ from $5S$ to obtain $4S= 1+\frac{1}{5^2}+\frac{1}{5^3}+\frac{2}{5^4}+...=1+\frac{1}{5}S$

Thus, $\frac{19}{5}S=1 \implies S= \frac{5}{19} =\frac {A}{B} \implies A+B=24$

All right.

For completeness, we should also mention that the series converges because the terms are $O\left(\frac{\phi^n}{5^n}\right)$ .

Brian Moehring - 4 years, 3 months ago

I asked for the sum as a fraction, isn't that enough to indicate that the sum converges?

Hana Wehbi - 4 years, 3 months ago

this is a Lehigh problem. the solution is wonderful though

Sathvik Acharya - 4 years, 3 months ago