# Does a_n=(-1/2)^n sequence converge or diverge? How do you find its limit?

## Does ${a}_{n} = {\left(- \frac{1}{2}\right)}^{n}$ sequence converge or diverge? How do you find its limit?

Dec 21, 2017

Sequence converges.

#### Explanation:

${a}_{n} = {\left(- \frac{1}{2}\right)}^{n}$

Let's look at a few terms of this sequence.

${a}_{1} = - \frac{1}{2}$

${a}_{2} = \frac{1}{4}$

${a}_{3} = - \frac{1}{8}$

This is a geometric progresion (GP) with first term ${a}_{1} = - \frac{1}{2}$ and common ratio $\left(r\right) = - \frac{1}{2}$

We are asked whether or not the sequence converges.

Consider, ${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} {\left(- \frac{1}{2}\right)}^{n} = 0$

$\therefore$ the sequence will converge.

Now let's consider the sum of the infinite series ${\sum}_{n = 1}^{\infty} {a}_{n}$

The sum of an infinite GP where $\left\mid r \right\mid < 0$ is given by ${a}_{1} / \left(1 - r\right)$

Hence, in our example:

sum_(n=1)^oo a_n = (-1/2)/(1-(-1/2)

$= - \frac{1}{2 \left(\frac{3}{2}\right)}$

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

So, we can state that the sequence converges and the sum of the infinite sequence converges to $- \frac{1}{3}$