Harmonic Mean and Infinite Sum

Algebra Level 4

∑ k = 1 ∞ 1 3 k a k \large \sum_{k=1}^\infty \frac{1}{3^ka_k}

A sequence { a n } \{a_n\} of real numbers is defined by a 1 = 20 , a 2 = 14 a_1=20,a_2=14 and for n ≥ 3 n\geq 3 , a n a_n is the harmonic mean of a n − 1 a_{n-1} and a n − 2 a_{n-2} .

If the infinite sum above can be represented in the form of p q \frac pq for some relatively prime positive integers p p and q q , find p + q p+q .


The answer is 403.

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Russelle Guadalupe by Russelle Guadalupe , 25, Philippines

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3 solutions

Patrick Corn
May 27, 2015

Let S S be the sum. Then S = 1 3 a 1 + 1 9 a 2 + 1 2 ∑ k = 3 ∞ ( 1 3 k a k − 1 + 1 3 k a k − 2 ) = 1 3 a 1 + 1 9 a 2 + 1 6 ∑ b = 2 ∞ 1 3 b a b + 1 18 ∑ c = 1 ∞ 1 3 c a c \begin{aligned} S &= \frac1{3a_1} + \frac1{9a_2} + \frac12 \sum_{k=3}^{\infty} \left( \frac1{3^k a_{k-1}} + \frac1{3^k a_{k-2}} \right) \\ &= \frac1{3a_1} + \frac1{9a_2} + \frac16 \sum_{b=2}^{\infty} \frac1{3^b a_b} + \frac1{18} \sum_{c=1}^{\infty} \frac1{3^c a_c} \end{aligned} where b = k − 1 b = k-1 and c = k − 2 c = k-2 .

So S = 1 3 a 1 + 1 9 a 2 + 1 6 ( S − 1 3 a 1 ) + 1 18 S S = \frac1{3a_1} + \frac1{9a_2} + \frac16 \left( S - \frac1{3a_1} \right) + \frac1{18} S and solving for S S leads to S = 45 126 a 1 + 1 7 a 2 S = \frac{45}{126a_1} + \frac1{7a_2} .

Plugging in a 1 = 20 , a 2 = 14 a_1 = 20, a_2 = 14 gives S = 11 392 S = \frac{11}{392} , so the answer is 403 \fbox{403} .

6 Helpful 0 Interesting 0 Brilliant 1 Confused
Aareyan Manzoor
Jun 16, 2015

2 Helpful 0 Interesting 1 Brilliant 0 Confused

Do not mind the "r≠1" part.

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