# Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.04 Geometric Series Example

## Example

Determine if the following series is convergent, and if so, find its sum:

{\begin{aligned}\sum _{k=0}^{\infty }(-1)^{k}{\frac {2^{k}}{3^{k}}}.\end{aligned}} ## Complete Solution

{\begin{aligned}\sum _{k=0}^{\infty }(-1)^{k}{\Big (}{\frac {2^{k}}{3^{k}}}{\Big )}&=\sum _{k=0}^{\infty }{\big (}{\frac {-2}{3}}{\big )}^{k}&&(1)\\&=\sum _{k=1}^{\infty }{\big (}{\frac {-2}{3}}{\big )}^{k-1}&&(2)\\&={\frac {1}{1-(-2/3)}}&&(3)\\&={\frac {3}{5}}\end{aligned}} The sum of the series is therefore 3/5.

## Explanation of Each Step

### Step (1)

We first rewrite the problem so that the summation starts at one and is in the familiar form of a geometric series, whose general form is

$\sum _{k=1}^{\infty }ar^{k-1}.$ After bringing the negative one and the three fifths together, we see that our given infinite series is geometric with common ratio -3/5.

For a geometric series to be convergent, its common ratio must be between -1 and +1, which it is, and so our infinite series is convergent.

We must now compute its sum.

### Step (2)

The given series

$\sum _{k=0}^{\infty }(-1)^{k}{\Big (}{\frac {2^{k}}{3^{k}}}{\Big )}=\sum _{k=0}^{\infty }{\big (}{\frac {-2}{3}}{\big )}^{k}$ starts the summation at $k=0$ , so we shift the index of summation by one:

$\sum _{k=0}^{\infty }{\big (}{\frac {-2}{3}}{\big )}^{k}=\sum _{k=1}^{\infty }{\big (}{\frac {-2}{3}}{\big )}^{k-1}.$ Our sum is now in the form of a geometric series with a = 1, r = -2/3. Since |r| < 1, the series converges, and its sum is

${\frac {a}{1-r}}={\frac {1}{1-(-2/3)}}={\frac {3}{5}}.$ ### Step (3)

In Step (3) we applied the formula for the sum of a geometric series:

{\begin{aligned}\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}\end{aligned}} This formula was derived in a previous section of this lesson.

## Possible Challenge Areas

### Step (2)

Students who may have been confused by this step may wish to refer to the previous lesson on Sigma Notation, where this process was explained in more detail.