Question

Find the values of x for which the series $\sum _{k=0}^{\infty }{\left(\frac{3}{x}\right)}^k$converges.

Short answer

The series $\sum _{k=0}^{\infty }{\left(\frac{3}{x}\right)}^k$ converges only for $x>3$or $x<-3$.

Step-by-step solution

Step 1. Given information.

Given a series $\sum _{k=0}^{\infty }{\left(\frac{3}{x}\right)}^k$.

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 series $\sum _{k=0}^{\infty }{\left(\frac{3}{x}\right)}^k$has $r=\frac{3}{x}$
.

For the series to converge, $\left|\frac{3}{x}\right|<1$.

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

It follows that localid="1648831728426" $\sum _{k=0}^{\infty }{\left(\frac{3}{x}\right)}^k$converges for all localid="1648831751828" $x>3$or localid="1648831755416" $x<-3$.