# Question #41dc8

Nov 14, 2017

See below.

#### Explanation:

This is a geometric series. The sum to infinity of a geometric series can be expressed as:

$a \left(\frac{1 - {r}^{n}}{1 - r}\right)$

Where $a$ is the first term, $r$ is the common ratio and n is the nth term.

The limit to infinity of:

${\lim}_{n \to \infty} \left(a - {r}^{n}\right) = a$ if and only if $- 1 < r < 1$

( since ${r}^{n} \to 0$, this is convergence)

If $\textcolor{w h i t e}{88} 1 < r < - 1$

Then:

$r > 1 \textcolor{w h i t e}{88}$

${\lim}_{n \to \infty} \left(a - {r}^{n}\right) = - \infty$ ( this is divergence )

( for $r < - 1$ the limit is undefined )

So from example:

Common difference is:

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

$- 1 < \frac{- \sqrt{3}}{3} < 1$ ( so this is a convergent series )

Sum to infinity:

$- \frac{1}{\sqrt{3}} \left(\frac{1}{1 - \frac{- \sqrt{3}}{3}}\right) = \frac{\frac{- 1}{\sqrt{3}}}{1 + \frac{\sqrt{3}}{3}} = - \frac{\sqrt{3}}{3 + \sqrt{3}}$