Serious series?

Number Theory Level 2

{ 1 3 , 1 15 , 1 35 , ⋯ 1 1599 } \large \left \{ \frac 13, \ \frac1{15}, \ \frac1{35}, \ \cdots \ \frac1{1599} \right \} Find the sum of the above series.


The answer is 0.4878.

Abhishek Alva by abhishek alva , 19, India

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2 solutions

Chew-Seong Cheong
May 30, 2016

S = 1 3 + 1 15 + 1 35 + ⋯ + 1 1599 = 1 1 ⋅ 3 + 1 3 ⋅ 5 + 1 5 ⋅ 7 + ⋯ + 1 39 ⋅ 41 = ∑ n = 1 20 1 ( 2 n − 1 ) ( 2 n + 1 ) = 1 2 ∑ n = 1 20 ( 1 2 n − 1 − 1 2 n + 1 ) = 1 2 ( ∑ n = 1 20 1 2 n − 1 − ∑ n = 2 21 1 2 n − 1 ) = 1 2 ( 1 − 1 41 ) = 20 41 ≈ 0.4878 \begin{aligned} S & = \frac 13 + \frac 1{15} + \frac 1{35} + \cdots + \frac 1{1599} \\ & = \frac 1{1 \cdot 3} + \frac 1{3\cdot 5} + \frac 1{5 \cdot 7} + \cdots + \frac 1{39 \cdot 41} \\ & = \sum_{n=1}^{20} \frac 1{(2n-1)(2n+1)} \\ & = \frac{1}{2} \sum_{n=1}^{20} \left( \frac 1{2n-1} - \frac 1{2n+1} \right) \\ & = \frac{1}{2} \left( \sum_{n=1}^{20} \frac 1{2n-1} - \sum_{n=2}^{21} \frac 1{2n-1} \right) \\ & = \frac{1}{2} \left( 1 - \frac 1{41} \right) \\ & = \frac{20}{41} \approx \boxed{0.4878} \end{aligned}

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Abhishek Alva
May 30, 2016

the above series cab be rewritten as 1/1 3 +1/3 5 +1/5 7........ +1/39 41 this is a telescopic series 1/2(1/1-1/3+1/3-1/5.......-1/41) now the middle fraction will cancel out. what remains is 1/2(1-1/41) =20/41=0.4878

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good solution

Ayush G Rai - 5 years ago

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thanks how ever its my problem

abhishek alva - 5 years ago

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Its easy to make this problem.

Ayush G Rai - 5 years ago

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