n-o-oo-x-n-4n-2-1-

Question Number 184768 by SANOGO last updated on 11/Jan/23

\underset{n = o}{\sum^{+ o o}} \frac{x^{n}}{4 n^{2} - 1}

Commented by SANOGO last updated on 11/Jan/23

m e r c i

Answered by mr W last updated on 11/Jan/23

\underset{n = 0}{\sum^{\infty}} x^{2 n} = 1 + x^{2} + x^{4} + \dots = \frac{1}{1 - x^{2}}

\underset{n = 0}{\sum^{\infty}} \int_{0}^{x} x^{2 n} d x = \int_{0}^{x} \frac{d x}{1 - x^{2}}

\underset{n = 0}{\sum^{\infty}} \frac{x^{2 n + 1}}{2 n + 1} = \frac{1}{2} \ln \frac{1 + x}{1 - x}

\underset{n = 0}{\sum^{\infty}} \frac{x^{2 n}}{2 n + 1} = \frac{1}{2 x} \ln \frac{1 + x}{1 - x}

r e p l a c e x w i t h \sqrt{x}

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{2 n + 1} = \frac{1}{2 \sqrt{x}} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}}

\underset{n = 1}{\sum^{\infty}} \frac{x^{n - 1}}{2 n - 1} = \frac{1}{2 \sqrt{x}} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}}

\underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{2 n - 1} = \frac{x}{2 \sqrt{x}} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}}

\underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{2 n - 1} + 1 = \frac{\sqrt{x}}{2} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}}

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{2 n - 1} = \frac{\sqrt{x}}{2} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}} - 1

\underset{n = 0}{\sum^{\infty}} (\frac{x^{n}}{2 n - 1} - \frac{x^{n}}{2 n + 1}) = \frac{\sqrt{x}}{2} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}} - 1 - \frac{1}{2 \sqrt{x}} \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}}

\Rightarrow 2 \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{4 n^{2} - 1} = \frac{1}{2} (\sqrt{x} - \frac{1}{\sqrt{x}}) \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}} - 1

\Rightarrow \underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{4 n^{2} - 1} = - \frac{1}{4} (\frac{1}{\sqrt{x}} - \sqrt{x}) \ln \frac{1 + \sqrt{x}}{1 - \sqrt{x}} - \frac{1}{2}

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