Unusual infinite series

Calculus Level 3

If S = 1 + 2 7 + 2 ⋅ 5 7 ⋅ 9 + 2 ⋅ 5 ⋅ 2 7 ⋅ 9 ⋅ 7 + 2 ⋅ 5 ⋅ 2 ⋅ 5 7 ⋅ 9 ⋅ 7 ⋅ 9 + 2 ⋅ 5 ⋅ 2 ⋅ 5 ⋅ 2 7 ⋅ 9 ⋅ 7 ⋅ 9 ⋅ 7 ⋯ S=1+\dfrac{2}{7}+\dfrac{2\cdot5}{7\cdot9}+\dfrac{2\cdot5\cdot2}{7\cdot9\cdot7}+\dfrac{2\cdot5\cdot2\cdot5}{7\cdot9\cdot7\cdot9}+\dfrac{2\cdot5\cdot2\cdot5\cdot2}{7\cdot9\cdot7\cdot9\cdot7}\cdots

And S can be represented by the form S = a b S=\dfrac{a}{b} for positive coprime integers a,b. Find a+b

This is just a quick filler problem. I'll be comming out with a set of my top ≈ e π − π 2 \approx \cfrac{e^{\pi}-\pi}{2} best problems soon. They are mostly challenging but all have very beautiful answers and computations


The answer is 134.

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Trevor Arashiro by Trevor Arashiro , 22, USA

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3 solutions

Trevor Arashiro
Mar 11, 2015

Regrouping

If S = ( 2 7 + 2 ⋅ 5 ⋅ 2 7 ⋅ 9 ⋅ 7 + 2 ⋅ 5 ⋅ 2 ⋅ 5 ⋅ 2 7 ⋅ 9 ⋅ 7 ⋅ 9 ⋅ 7 ⋯ ) + ( 1 + 2 ⋅ 5 7 ⋅ 9 + 2 ⋅ 5 ⋅ 2 ⋅ 5 7 ⋅ 9 ⋅ 7 ⋅ 9 + 2 ⋅ 5 ⋅ 2 ⋅ 5 ⋅ 2 ⋅ 5 7 ⋅ 9 ⋅ 7 ⋅ 9 ⋅ 7 ⋅ 9 ⋯ ) S=\left(\dfrac{2}{7}+\dfrac{2\cdot5\cdot2}{7\cdot9\cdot7}+\dfrac{2\cdot5\cdot2\cdot5\cdot2}{7\cdot9\cdot7\cdot9\cdot7}\cdots \right)+\left(1+\dfrac{2\cdot5}{7\cdot9}+\dfrac{2\cdot5\cdot2\cdot5}{7\cdot9\cdot7\cdot9}+\dfrac{2\cdot5\cdot2\cdot5\cdot2\cdot5}{7\cdot9\cdot7\cdot9\cdot7\cdot9}\cdots\right)

S = 2 7 ( 1 + 5 ⋅ 2 9 ⋅ 7 + 5 ⋅ 2 ⋅ 5 ⋅ 2 9 ⋅ 7 ⋅ 9 ⋅ 7 ⋯ ) + ( 1 + 2 ⋅ 5 7 ⋅ 9 + 2 ⋅ 5 ⋅ 2 ⋅ 5 7 ⋅ 9 ⋅ 7 ⋅ 9 + 2 ⋅ 5 ⋅ 2 ⋅ 5 ⋅ 2 ⋅ 5 7 ⋅ 9 ⋅ 7 ⋅ 9 ⋅ 7 ⋅ 9 ⋯ ) S=\dfrac{2}{7}\left(1+\dfrac{5\cdot2}{9\cdot7}+\dfrac{5\cdot2\cdot5\cdot2}{9\cdot7\cdot9\cdot7}\cdots \right)+\left(1+\dfrac{2\cdot5}{7\cdot9}+\dfrac{2\cdot5\cdot2\cdot5}{7\cdot9\cdot7\cdot9}+\dfrac{2\cdot5\cdot2\cdot5\cdot2\cdot5}{7\cdot9\cdot7\cdot9\cdot7\cdot9}\cdots\right)

S = 2 7 ( ∑ n = 0 ∞ ( 2 ⋅ 5 7 ⋅ 9 ) n ) + ( ∑ n = 0 ∞ ( 2 ⋅ 5 7 ⋅ 9 ) n ) S=\dfrac{2}{7}\left(\displaystyle \sum_{n=0}^{\infty}\left(\dfrac{2\cdot5}{7\cdot9}\right)^n\right)+\left(\displaystyle \sum_{n=0}^{\infty}\left(\dfrac{2\cdot5}{7\cdot9}\right)^n\right)

S = 9 7 ( ∑ n = 0 ∞ ( 2 ⋅ 5 7 ⋅ 9 ) n ) S=\dfrac{9}{7}\left(\displaystyle \sum_{n=0}^{\infty}\left(\dfrac{2\cdot5}{7\cdot9}\right)^n\right)

S = 9 7 ( 1 1 − 10 63 ) S=\dfrac{9}{7}\left(\dfrac{1}{1-\frac{10}{63}}\right)

S = 81 53 S=\dfrac{81}{53}

Therefore, a + b = 134 a+b=134

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I hadn't realized just how close e π − π e^{\pi} - \pi is to 20. 20. I mean, 19.999... 19.999... is kind of freaky. :) That brings to mind another question: without a calculator, determine which of ( e π − π ) (e^{\pi} - \pi) or ( π e − e ) (\pi^{e} - e) is the greater.

Anyway, nice little filler question until your top ~10; looking forward to seeing what you have in store for us.

Brian Charlesworth - 6 years, 3 months ago

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They are for the most part quite hard. I remember you saying that it was near when the golden ratio popped up in questions so I tried to incorporate it as much as I possibly could :). I think it pops up about 4 times.

Trevor Arashiro - 6 years, 3 months ago

did the same way..

Tanishq Varshney - 6 years, 3 months ago

Let the sum be equal to S . Multiply S By 10/63 . subtract it from S. All terms will cancel out except 1+2/7. So, extracting S on one side we get S =(9/7)/(53/63)=81/53

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Same method !!

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