SeriesMedium

Power series: radius of convergence

Find the radius of convergence of

\displaystyle\sum_{n=1}^\infty \dfrac{n\,x^n}{3^n}

Given data

c_n = n/3^n

Steps

0 / 3

Apply the ratio test: calculate |a_{n+1}/a_n|.
Write your working — the numerical result must appear in the text (logica)
Calculate the limit as n→∞.
The series converges for |x|/3 < 1. Find R.

Full worked solution

Apply the ratio test: calculate |a_{n+1}/a_n|.
$\left|\frac{(n+1)x^{n+1}/3^{n+1}}{nx^n/3^n}\right| = \frac{(n+1)|x|}{3n}$
$\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{(n+1)x^{n+1}/3^{n+1}}{n x^n/3^n}\right| = \frac{n+1}{n} \cdot \frac{|x|}{3}$ . The ratio of consecutive terms simplifies by cancelling powers of $x$ and $3$ .
Calculate the limit as n→∞.
$\lim_{n\to\infty}\frac{(n+1)|x|}{3n} = \frac{|x|}{3}$
$\lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty} \frac{n+1}{n} \cdot \frac{|x|}{3} = 1 \cdot \frac{|x|}{3} = \frac{|x|}{3}$ . As $n\to\infty$ , $(n+1)/n \to 1$ .
The series converges for |x|/3 < 1. Find R.
$R = 3$
The ratio test requires $\frac{|x|}{3} < 1 \rightarrow |x| < 3 \rightarrow \mathbf{R = 3}$ . The radius of convergence is 3.

Result:R = 3.

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