Answer to Question #142661 in Calculus for Moel Tariburu

Question

Find the radius of convergence for the series:
∑_(k=1)^∞▒〖-5x〗^k/(k+2)

Answer 1

Use ratio test as,

$L$$=\lim\nolimits_x\rarr\infin∣ \frac{a_{k+1}}{a_{k}}∣$

$=\lim\nolimits_x\rarr\infin∣ \frac{(-5x)^{k+1}}{(k+1)+2}\frac{k+2}{(-5x)^{k}}∣$

$=\lim\nolimits_x\rarr\infin∣ \frac{(-5x)^{k+1}}{(-5x)^{k}}\frac{k+2}{k+3}∣$

$=\lim\nolimits_x\rarr\infin∣ (-5x)\frac{k(1+\frac{2}{k})}{k(1+\frac{3}{k})}∣$

$=∣ (-5x)\frac{(1+0)}{(1+0)}∣$

$=∣ (-5x)∣$

$=∣ 5x∣$

For convergence $∣ 5x∣<1$ , so

$-1<5x<1$

$-\frac{1}{5}<x<\frac{1}{5}$

The radius of convergence of the series is,

$R=\frac{1}{2}[\frac{1}{5}-(-\frac{1}{5})]$

$=\frac{1}{2}[\frac{1}{5}+\frac{1}{5}]$

$=\frac{1}{2}[\frac{2}{5}]$

$=\frac{1}{5}$

Therefore, the radius of convergence is $R=\frac{1}{5}$