Question

Find the values of x for which the series$\sum _{k=0}^{\infty }{x}^k$converges.

Short answer

The series $\sum _{k=0}^{\infty }{x}^k$converges only for$-1<x<1$.

Step-by-step solution

Step 1. Given information.

Given a series$\sum _{k=0}^{\infty }{x}^k$.

Step 2. Find all values of x for which the series converges.

A geometric series is of the form $\sum _{k=0}^{\infty }c{r}^k$ for some constants c and r.

Supposer is a non-zero real number, then $\sum _{k=0}^{\infty }c{r}^k$converges to $\frac{c}{1-r}$if and only if $\left|r\right|<1$.

Here, the serieslocalid="1648830940128" $\sum _{k=0}^{\infty }{x}^k$has $r=x$.

For the series to converge, $\left|x\right|<1$.

Note that if$\left|y\right|<a$, it follows that$-a<y<a$.

It follows that $\sum _{k=0}^{\infty }{x}^k$converges for all $-1<x<1$.