infinite series - 6

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \cdots\)

 

[logistic_guy]

\(\displaystyle \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots\)

 

[logistic_guy]

\(\displaystyle \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \sum_{n=1}^{\infty}\frac{1}{(2n - 1)(2n + 1)}\)

 

[logistic_guy]

Let us do partial fraction decomposition.

\(\displaystyle \frac{1}{(2n - 1)(2n + 1)} = \frac{A}{2n - 1} + \frac{B}{2n + 1}\)


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


\(\displaystyle A = \frac{1}{2}\)

\(\displaystyle B = -\frac{1}{2}\)

Then,

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n - 1)(2n + 1)} = \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)\)

 

[logistic_guy]

\(\displaystyle \frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right) = \lim_{N \rightarrow \infty}\frac{1}{2}\sum_{n=1}^{N}\left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)\)


\(\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2}\left[\left(\frac{1}{1}\right) - \left(\frac{1}{3}\right) + \left(\frac{1}{3}\right) - \left(\frac{1}{5}\right) + \left(\frac{1}{5}\right) - \left(\frac{1}{7}\right) + \cdots + \left(\frac{1}{2N - 1}\right) - \left(\frac{1}{2N + 1}\right)\right]\)


\(\displaystyle = \lim_{N \rightarrow \infty}\frac{1}{2}\left[\left(\frac{1}{1}\right) - \left(\frac{1}{2N + 1}\right)\right] = \frac{1}{2}\left[\left(\frac{1}{1}\right)\right] = \frac{1}{2}\)

Then,

\(\displaystyle \frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \cdots = \textcolor{blue}{\frac{1}{2}}\)