# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1/(1*3)+1/(3*5)+1/(5*7)+...+1/((2n-1)(2n+1))+...?

Mar 19, 2017

The nth partial sunm is $= \frac{n}{2 n + 1}$
The series converge to $\frac{1}{2}$

#### Explanation:

Let's perform the partial fraction decomposition of the nth term

$\frac{1}{\left(2 n - 1\right) \left(2 n + 1\right)} = \frac{A}{2 n - 1} + \frac{B}{2 n + 1}$

$= \frac{A \left(2 n + 1\right) + B \left(2 n - 1\right)}{\left(2 n - 1\right) \left(2 n + 1\right)}$

The numerators are the same, we compare the denominators

1=A(2n+1)+B(2n-1))

Let $n = \frac{1}{2}$, $\implies$, $1 = 2 A$, $\implies$, $A = \frac{1}{2}$

Let $n = - \frac{1}{2}$, $\implies$, $1 = - 2 B$, $\implies$, $B = - \frac{1}{2}$

Therefore,

$\frac{1}{\left(2 n - 1\right) \left(2 n + 1\right)} = \frac{\frac{1}{2}}{2 n - 1} - \frac{\frac{1}{2}}{2 n + 1}$

${u}_{n} = \frac{1}{2} \left(\frac{1}{2 n - 1} - \frac{1}{2 n + 1}\right)$

So,

${u}_{1} = \frac{1}{2} \left(\frac{1}{1} - \cancel{\frac{1}{3}}\right)$

${u}_{2} = \frac{1}{2} \left(\cancel{\frac{1}{3}} - \cancel{\frac{1}{5}}\right)$

${u}_{3} = \frac{1}{2} \left(\cancel{\frac{1}{5}} - \cancel{\frac{1}{7}}\right)$

${u}_{n - 1} = \frac{1}{2} \left(\cancel{\frac{1}{2 n - 3}} - \cancel{\frac{1}{2 n - 1}}\right)$

${u}_{n} = \frac{1}{2} \left(\cancel{\frac{1}{2 n - 1}} - \frac{1}{2 n + 1}\right)$

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

$= \frac{n}{2 n + 1}$

The nth partial sunm is $= \frac{n}{2 n + 1}$

${\lim}_{n \to + \infty} \frac{n}{2 n + 1} = {\lim}_{n \to + \infty} \frac{n}{2 n} = \frac{1}{2}$

The series converge to $\frac{1}{2}$