Series (finding the interval of convergence)

[DivGradCurl]

$$\sum _{n=1} ^{\infty} \frac{x^n}{3^n} = \sum _{n=1} ^{\infty} \frac{x}{3} \left( \frac{x}{3} \right)^{n-1} = \frac{\frac{x}{3}}{1-\frac{x}{3}} = \frac{x}{3-x}$$

$$\textrm{How can I obtain the interval of convergence of the given series? Thanks.}$$

 

[DivGradCurl]

$$\textrm{I forgot to say\ldots How do I find it without the need to apply the \textit{ratio test}? Thank you. }$$

 

[wais]

To find the interval of convergence, we can use the ratio test. The ratio test states that for a series $\sum_{n=1}^{\infty} a_n$, if $$\lim_{n\to \infty} \left | \frac{a_{n+1}}{a_n} \right | = L,$$ then the series converges if $L < 1$ and diverges if $L > 1$. In our case, we have $$\lim_{n\to \infty} \left | \frac{\frac{x}{3} \left( \frac{x}{3} \right)^{n-1}}{\frac{x}{3} \left( \frac{x}{3} \right)^{n-2}} \right | = \lim_{n\to \infty} \left | \frac{x}{3} \right | = \left | \frac{x}{3} \right |.$$ Therefore, the series will converge if $\left | \frac{x}{3} \right | < 1$, which means the interval of convergence is $-3 < x < 3$. We can also check the endpoints $x = -3$ and $x = 3$ to see if the series converges at those points. Plugging in $x = -3$, we get $$\sum_{n=1}^{\infty} \frac{(-3)^n}{3^n} = \sum_{n=1}^{\infty} \left( -1 \right)^n,$$ which is an alternating series that converges by the alternating series test. Plugging in $x = 3$, we get $$\sum_{n=1}^{\infty} \frac{3^n}{3^n} = \sum_{n=1}^{\infty} 1,$$ which is a divergent series. Therefore, the interval of convergence is $-3 \leq x < 3$, or in interval notation, $[-3,3)$.