# If sum_(n=2) ^oo (1+k)^-n=2 what is k?

## The answer is supposed to be $k = \frac{\sqrt{3} - 1}{2}$ I just don't know how to get there.

Dec 14, 2016

$k = \frac{\sqrt{3} - 1}{2}$

#### Explanation:

A geometric series of the form ${\sum}_{n = 0}^{\infty} {r}^{n}$ with $| r | < 1$ evaluates to

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

With that,

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

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

$= - 1 - \frac{1}{1 + k} + \frac{1}{1 - \left(\frac{1}{1 + k}\right)}$

$= - 1 - \frac{1}{1 + k} + \frac{1}{\frac{k}{1 + k}}$

$= - 1 - \frac{1}{1 + k} + \frac{1 + k}{k}$

$= 2$

We can now solve for $k$. Multiplying through by $k \left(1 + k\right)$, we get

$- k \left(1 + k\right) - k + {\left(1 + k\right)}^{2} = 2 k \left(1 + k\right)$

$\implies - k - {k}^{2} - k + 1 + 2 k + {k}^{2} = 2 {k}^{2} + 2 k$

$\implies 1 = 2 {k}^{2} + 2 k$

$\implies 2 {k}^{2} + 2 k - 1 = 0$

$\implies k = \frac{- 1 \pm \sqrt{3}}{2}$

But we must have $k > 0 \mathmr{and} k < - 2$ for ${\lim}_{n \to \infty} {\left(1 + k\right)}^{- n} = 0$, a necessary condition for convergence, thus our only possibility becomes the positive option:

$k = \frac{\sqrt{3} - 1}{2}$