Strategic Minimalism

Calculus Level 5

The value of the infinite sum

1 5 + 1 10 + 3 85 + 1 65 + 5 629 + 3 650 + 7 2405 + ( ⋯ ) \dfrac{1}{5} + \dfrac{1}{10} + \dfrac{3}{85} + \dfrac{1}{65} + \dfrac{5}{629} + \dfrac{3}{650} + \dfrac{7}{2405} + (\cdots)

can be written as a b \dfrac{a}{b} , where a a and b b are positive coprime integers. Evaluate a + b a+b .


The answer is 11.

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Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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1 solution

Pranjal Jain
Jan 11, 2015

Seeing it for a while, it seems like 1 5 + 1 10 + 3 85 + 1 65 + 5 629 + . . . = 1 5 + 2 20 + 3 85 + 4 260 + 5 629 + . . . \dfrac{1}{5}+\dfrac{1}{10}+\dfrac{3}{85}+\dfrac{1}{65}+\dfrac{5}{629}+...\\=\dfrac{1}{5}+\dfrac{2}{20}+\dfrac{3}{85}+\dfrac{4}{260}+\dfrac{5}{629}+...

T n = n n 4 + 4 = n ( n 2 + 2 ) 2 − 4 n 2 = n ( n 2 + 2 n + 2 ) ( n 2 − 2 n + 2 ) = 1 4 ( 1 n 2 − 2 n + 2 − 1 n 2 + 2 n + 2 ) T_{n}=\dfrac{n}{n^4+4}\\=\dfrac{n}{(n^2+2)^2-4n^2}\\=\dfrac{n}{(n^2+2n+2)(n^2-2n+2)}\\=\dfrac{1}{4}\left (\dfrac{1}{n^2-2n+2}-\dfrac{1}{n^2+2n+2}\right )

S = 1 4 ( 1 − 1 5 + 1 2 − 1 10 + 1 5 − 1 17 + 1 10 − 1 26 + . . . ) S=\dfrac{1}{4}\left (1-\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{1}{17}+\dfrac{1}{10}-\dfrac{1}{26}+...\right)

By telescoping technique, S = 3 8 S=\dfrac{3}{8}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Perfect reasoning, except for the end. Could you explain a little better the "telescoping technique"?

Guilherme Dela Corte - 6 years, 4 months ago

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Hi , I think that you should read this .

A Former Brilliant Member - 6 years, 4 months ago

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