Where's my telescope?

Algebra Level 3

$\Large \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{4n^2-1} ~=~?$

The answer is 0.5.

2 solutions

Rishabh Jain
Jan 10, 2016

Here's your telescope.... :) $\displaystyle \sum_{n=1}^∞\frac{1}{(2n+1)(2n-1)}$ $=\frac{1}{2}\displaystyle \sum_{n=1}^∞\frac{1}{2n-1}-\frac{1}{2n+1}\space\space \color{#D61F06}{(Telescopic\space series)}$ $=\frac{1}{2} \times 1 =\Large\color{#20A900}{1/2=0.5}$

same way exactly

Kaustubh Miglani - 5 years, 5 months ago

Why summation is equal to 1. If we substitute numbers it would be like this: 1+1/3+1/5+1/7+1/9....-1/5-1/7-1/9=1+1/3=4/3

Eziz Hudaykulyyev - 5 years, 5 months ago

You forgot -1/3 since at n=1

$-\frac{1}{2n+1}$ is -1/3

Rishabh Jain - 5 years, 5 months ago

Yeah thanks, I realized my fault.

Eziz Hudaykulyyev - 5 years, 5 months ago

Gabor Koranyi
Aug 25, 2018

Same as Rishabh Cool's.

Where's my telescope?

The answer is 0.5.

2 solutions

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