# Evaluate the taylor series sum: sum_{n=2}^{∞} {(-1)^n 3^{n-1}(2n)}/(2^{2n-1}) ?

## sum_{n=2}^{∞} {(-1)^n 3^{n-1}(2n)}/(2^{2n-1}) use taylor series to evaluate this sum.

Jun 22, 2018

${\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{\left(2 n\right) {3}^{n - 1}}{2} ^ \left(2 n - 1\right) = \frac{33}{49}$

#### Explanation:

Start from the geometric series:

${\sum}_{n = 0}^{\infty} {q}^{n} = \frac{1}{1 - q}$ for $\left\mid q \right\mid < 1$

Let $q = - x$ to have:

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

Differentiate term by term:

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

Let $x = \frac{3}{4}$:

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

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} \frac{n {3}^{n - 1}}{2} ^ \left(2 n - 2\right) = - \frac{16}{49}$

${\sum}_{n = 1}^{\infty} {\left(- 1\right)}^{n} \frac{\left(2 n\right) {3}^{n - 1}}{2} ^ \left(2 n - 1\right) = - \frac{16}{49}$

Extract the term for $n = 1$:

$- 1 + {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{\left(2 n\right) {3}^{n - 1}}{2} ^ \left(2 n - 1\right) = - \frac{16}{49}$

${\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{\left(2 n\right) {3}^{n - 1}}{2} ^ \left(2 n - 1\right) = - \frac{16}{49} + 1$

${\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} \frac{\left(2 n\right) {3}^{n - 1}}{2} ^ \left(2 n - 1\right) = \frac{33}{49}$