# How do you find the sum of the infinite geometric series Sigma -10(0.2)^n from n=0 to oo?

Dec 29, 2017

Use the formula for the sum of an infinite geometric series to obtain an answer of $- \frac{25}{2} = - 12.5$.

#### Explanation:

The formula can be written as ${\sum}_{n = 0}^{\infty} a \cdot {r}^{n} = \frac{a}{1 - r}$ whenever $| r | < 1$.

In the present example, $a = - 10$ and $r = 0.2$ so the answer is $\frac{- 10}{1 - 0.2} = - \frac{10}{0.8} = - 10 \cdot \frac{5}{4} = - \frac{25}{2} = - 12.5$.

To derive the formula ${\sum}_{n = 0}^{\infty} a \cdot {r}^{n} = \frac{a}{1 - r}$, consider the ${k}^{t h}$ partial sum ${s}_{k} = {\sum}_{n = 0}^{k} a \cdot {r}^{n}$. Multiplying both sides of this equation by $r$ gives $r {s}_{k} = {\sum}_{n = 0}^{k} a \cdot {r}^{n + 1} = {\sum}_{n = 1}^{k + 1} a \cdot {r}^{n}$.

Therefore, $\left(1 - r\right) {s}_{k} = {s}_{k} - r {s}_{k} = a - a {r}^{k + 1}$, implying that, when $r \ne 1$, ${s}_{k} = \frac{a \left(1 - {r}^{k + 1}\right)}{1 - r}$. (When $r = 1$, then ${s}_{k} = \left(k + 1\right) a$.)

If $| r | < 1$, then ${r}^{k + 1} \to 0$ as $k \to \infty$. This leads to the conclusion that

${\sum}_{n = 0}^{\infty} a \cdot {r}^{n} = {\lim}_{k \to \infty} {s}_{k} = \frac{a \left(1 - 0\right)}{1 - r} = \frac{a}{1 - r}$ when $| r | < 1$.

When $| r | \ge 1$, then the sequence $\left({r}^{k + 1}\right)$ diverges, making $\left({s}_{k}\right)$ diverge, leading to the conclusion that the series ${\sum}_{n = 0}^{\infty} a \cdot {r}^{n}$ diverges.