tinfoQ

2021-10-09

Prove that for every n in the set of natural numbers, N:
$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{\left(2n-1\right)\cdot \left(2n+1\right)}=\frac{n}{2n+1}$

Latisha Oneil

Step 1
The partial fraction is the process of splitting the terms in the denominator into different fractions. This can be used to simplify the fraction for further evaluation.
Step 2
The series is evaluated by splitting the terms into partial fractions as follows:-
$\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\dots +\frac{1}{\left(2n-1\right)\cdot \left(2n+1\right)}=\left(\frac{\frac{1}{2}}{1}-\frac{\frac{1}{2}}{3}\right)+\left(\frac{\frac{1}{2}}{3}-\frac{\frac{1}{2}}{5}\right)+\left(\frac{\frac{1}{2}}{5}-\frac{\frac{1}{2}}{7}\right)+\dots \left(\frac{\frac{1}{2}}{\left(2n-1\right)}-\frac{\frac{1}{2}}{\left(2n+1\right)}\right)$
$=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\dots +\frac{1}{2n-1}-\frac{1}{2n+1}\right)$
$=\frac{1}{2}\left(1-\frac{1}{2n+1}\right)$
$=\frac{1}{2}\left(\frac{2n+1-1}{2n+1}\right)$
$=\frac{1}{2}\left(\frac{2n}{2n+1}\right)$
$=\frac{n}{2n+1}$
Thus the sum of the series is obtained to be $\frac{n}{2n+1}$

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