What is the value of the sum 1/1*3 + 1/3*5 + 1/5*7+ 1/7*9+...+1/199*201? Express your answer as a fraction in simplest form.

What is the value of the sum 1/1*3 + 1/3*5 + 1/5*7+ 1/7*9+...+1/199*201?

Express your answer as a fraction in simplest form.

$\begin{array}{rcll} && \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7}+ \frac{1}{7*9}+\ldots+\frac{1}{199*201} \\ &=& \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7}+ \frac{1}{7*9}+\ldots+\frac{1}{(2n-1)(2n+1)} \\ \hline && \frac{1}{(2n-1)(2n+1)} = \frac12\left( \frac{1}{2n-1} - \frac{1}{2n+1} \right) \\ && \frac{1}{1*3} = \frac12\left( \frac{1}{1} - \frac{1}{3} \right) \\ && \frac{1}{3*5} = \frac12\left( \frac{1}{3} - \frac{1}{5} \right) \\ && \frac{1}{5*7} = \frac12\left( \frac{1}{5} - \frac{1}{7} \right) \\ && \frac{1}{7*9} = \frac12\left( \frac{1}{7} - \frac{1}{9} \right) \\ && \ldots \\ && \frac{1}{199*201} = \frac12\left( \frac{1}{199} - \frac{1}{201} \right) \\ \hline &=& \frac12\left( \frac{1}{1} - \frac{1}{3} \right) + \frac12\left( \frac{1}{3} - \frac{1}{5} \right) + \frac12\left( \frac{1}{5} - \frac{1}{7} \right) + \frac12\left( \frac{1}{7} - \frac{1}{9} \right)+\ldots+\frac12\left( \frac{1}{199} - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{1}{1} - \underbrace{\frac{1}{3} + \frac{1}{3}}_{=0} - \underbrace{\frac{1}{5}+\frac{1}{5}}_{=0} - \underbrace{\frac{1}{7} + \frac{1}{7}}_{=0} - \underbrace{\frac{1}{9}+ \frac{1}{9}}_{=0} +\ldots- \underbrace{\frac{1}{199}+\frac{1}{199}}_{=0} - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{1}{1} - \frac{1}{201} \right) \\ &=& \frac12\left( 1 - \frac{1}{201} \right) \\ &=& \frac12\left( \frac{201-1}{201} \right) \\ &=& \frac12\left( \frac{200}{201} \right) \\ &=& \frac{100}{201} \\ \end{array}$

∑[1/((2 n + 1) (2 n + 3)),n, 0, 201] =202/405 =~converges to 1/2.

I think that's incorrect