# Find the value of the taylor series sum: sum_{n=4}^∞ \frac{(n+1)(n)2^n}{3^n} ?

## sum_{n=4}^∞ \frac{(n+1)(n)2^n}{3^n} (find value of taylor series sum)

Jun 7, 2018

${\sum}_{n = 4}^{\infty} \frac{n \left(n + 1\right) {2}^{n}}{3} ^ n = \frac{256}{9}$

#### Explanation:

Start from the geometric series:

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

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

Multiply by $x$:

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

Differentiate term by term:

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

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

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

and again:

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

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

Let $x = \frac{2}{3}$. as $0 < \frac{2}{3} < 1$ this point is in the interval of convergence, so:

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

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

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

$36 = {\sum}_{n = 1}^{\infty} \frac{n \left(n + 1\right) {2}^{n}}{3} ^ n$

Extract now the terms for $n = 1$, $n = 2$ and $n = 3$ from the sum:

$36 = \frac{4}{3} + \frac{24}{9} + \frac{96}{27} + {\sum}_{n = 4}^{\infty} \frac{n \left(n + 1\right) {2}^{n}}{3} ^ n$

$36 - - \frac{4}{3} - \frac{24}{9} - \frac{96}{27} = {\sum}_{n = 4}^{\infty} \frac{n \left(n + 1\right) {2}^{n}}{3} ^ n$

$36 - \frac{12 + 24 + 32}{9} = {\sum}_{n = 4}^{\infty} \frac{n \left(n + 1\right) {2}^{n}}{3} ^ n$

$\frac{256}{9} = {\sum}_{n = 4}^{\infty} \frac{n \left(n + 1\right) {2}^{n}}{3} ^ n$