Question

How do you find the interval of convergence for a geometric series $?$

Answer 1

Interval of convergence for a geometric series:

The interval on which the series converges is called the interval of convergence.

Lets find the interval of convergence for the following geometric series:

$\sum _{\mathrm{n}=0}^{\infty }{\left(\frac{\mathrm{x}}{2}\right)}^{\mathrm{n}}=1+{\left(\frac{\mathrm{x}}{2}\right)}^1+{\left(\frac{\mathrm{x}}{2}\right)}^2+{\left(\frac{\mathrm{x}}{2}\right)}^3........\infty$

The common ratio $\left(\mathrm{r}\right)$ of the geometric series is:

$\mathrm{r}=\frac{\mathrm{x}}{2}$

If the above geometric series converges then:

$$\begin{gathered} \left|r\right|<1 \\ \Rightarrow -1<\frac{\mathrm{x}}{2}<1 \\ \Rightarrow -2<\mathrm{x}<2 \end{gathered}$$

So, the interval of convergence of the geometric series will be $\left(-2,2\right)$.

In this way, the interval of convergence of a geometric series can be found.

Therefore, determine if the power series will converge for $x=a-R\mathrm{or}x=a+R$.

If the series converges for any of these values then it will include in the interval of convergence.