# Find the value of the power series sum: sum_{n=2}^∞ \frac{(-1)^n}{(2n+1)(2n+2)3^n} ?

Jun 7, 2018

${\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{1}{3} ^ n \frac{1}{\left(2 n + 1\right) \left(2 n + 2\right)} = \sqrt{3} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{3}{2} \ln \left(\frac{4}{3}\right) - \frac{17}{36}$

#### Explanation:

Consider the geometric series:

${\sum}_{n = 0}^{\infty} {q}^{n} = \frac{1}{1 - q}$

for $\left\mid q \right\mid < 1$.

Let $q = - {x}^{2}$

$\frac{1}{1 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n}$

Integrate term by term:

${\int}_{0}^{x} \frac{\mathrm{dt}}{1 + {t}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\int}_{0}^{x} {t}^{2 n} \mathrm{dt}$

$\arctan x = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 1} / \left(2 n + 1\right) \mathrm{dt}$

Integrate again, considering that integrating by parts:

$\int \arctan x \mathrm{dx} = x \arctan x - \int \frac{x}{1 + {x}^{2}} \mathrm{dx}$

$\int \arctan x \mathrm{dx} = x \arctan x - \frac{1}{2} \int \frac{d \left(1 + {x}^{2}\right)}{1 + {x}^{2}}$

$\int \arctan x \mathrm{dx} = x \arctan x - \frac{1}{2} \ln \left(1 + {x}^{2}\right) + c$

so:

${\int}_{0}^{x} \arctan t = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right) {\int}_{0}^{x} {t}^{2 n + 1} \mathrm{dt}$

$x \arctan x - \frac{1}{2} \ln \left(1 + {x}^{2}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(\left(2 n + 1\right) \left(2 n + 2\right)\right)$

Let now $x = \frac{1}{\sqrt{3}}$. As $0 < \frac{1}{\sqrt{3}} < 1$ the series is convergent in this point:

$\frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln \left(1 + \frac{1}{3}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{1}{\sqrt{3}} ^ \left(2 n + 2\right) \frac{1}{\left(2 n + 1\right) \left(2 n + 2\right)}$

note that:

${\left(\sqrt{3}\right)}^{2 n + 2} = 3 \cdot {\left(\sqrt{3}\right)}^{2 n} = 3 \cdot {3}^{n}$

so:

$\frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln \left(1 + \frac{1}{3}\right) = \frac{1}{3} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \frac{1}{3} ^ n \frac{1}{\left(2 n + 1\right) \left(2 n + 2\right)}$

Extract now the first two terms for $n = 0$ and $n = 1$:

$\frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln \left(1 + \frac{1}{3}\right) = \frac{1}{6} - \frac{1}{108} + \frac{1}{3} {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{1}{3} ^ n \frac{1}{\left(2 n + 1\right) \left(2 n + 2\right)}$

$\frac{1}{\sqrt{3}} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{1}{2} \ln \left(1 + \frac{1}{3}\right) - \frac{1}{6} + \frac{1}{108} = \frac{1}{3} {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{1}{3} ^ n \frac{1}{\left(2 n + 1\right) \left(2 n + 2\right)}$

$\sqrt{3} \arctan \left(\frac{1}{\sqrt{3}}\right) - \frac{3}{2} \ln \left(\frac{4}{3}\right) - \frac{17}{36} = {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{1}{3} ^ n \frac{1}{\left(2 n + 1\right) \left(2 n + 2\right)}$