Question

Find the sum S below:

Answer 1

Answer:

$\rm\displaystyle \sum _{n = 1} ^{ \infty } \frac{2}{(2n -1 )(2n + 1)} \approx \boxed{\frac{1}{2}}$

Step-by-step explanation:

we want to figure out the Sum of the following infinite series:

$\rm\displaystyle S = \frac{2}{1 \cdot 3} + \frac{2}{3 \cdot 5} + \frac{2}{5 \cdot 7} + ... + \frac{2}{(2n - 1)(2n + 1)}$

we're given the nth term of the series so the infinite sum would be

$\displaystyle \sum _{n = 1} ^{ \infty } \frac{2}{(2n -1 )(2n + 1)}$

to simplify the sum it, we need our calculas skills. recall that,

$\displaystyle \sum _{i = 1} ^{ \infty } a_{i} = \lim _{x \to \infty }\sum _{i = 1} ^{ n } a_{i}$

Thus,

$\displaystyle \sum _{n = 1} ^{ \infty } \frac{2}{(2n -1 )(2n + 1)} = \lim _{x \to \infty } \sum _{n = 1} ^{ k} \frac{2}{(2n -1 )(2n + 1)}$

computing the limit:

$\displaystyle \lim _{n \to \infty } \frac{2}{ \dfrac{(2n -1 )(2n + 1)}{ {n}^{2} }}$

simplify complex fraction:

$\displaystyle \lim _{n \to \infty } \frac{ {2n}^{2} }{ {(2n -1 )(2n + 1)}}$

simplify denominator:

$\displaystyle \lim _{n \to \infty } \frac{ {2n}^{2} }{ { 4n}^{2} - 1}$

factor out n²:

$\displaystyle \lim _{n \to \infty } \frac{ {2n}^{2} }{ {n}^{2} \left( 4 - \frac{1}{ {n}^{2}} \right) }$

reduce fraction:

$\displaystyle \lim _{n \to \infty } \frac{ 2 }{ 4 - \dfrac{1}{ {n}^{2}} }$

as n approaches to infinity 1/n² approaches to 0 therefore:

$\displaystyle \frac{ 2 }{ 4 }$

reduce fraction:

$\displaystyle \frac{1}{2}$

hence,

$\displaystyle \sum _{n = 1} ^{ \infty } \frac{2}{(2n -1 )(2n + 1)} \approx \frac{1}{2}$