Answer to Question #259868 in Calculus for Dhanush

Question

Using weiestrass M test show that the following series converges uniformly sigma infinity n=1 n³ x^n x€[-1/3,1/3]

Answer 1

Using Weiestrass M test let us show that the following series converges uniformly $\sum_{n=1}^{ \infty} n^3 x^n,\ x\in[-\frac{1}{3},\frac{1}{3}].$

Taking into account that

$\max\limits_{x\in[-\frac{1}{3},\frac{1}{3}]}{|n^3 x^n|}=\frac{n^3}{3^n},$ and

$\lim\limits_{n\to\infty}\frac{\frac{(n+1)^3}{3^{n+1}}}{\frac{n^3}{3^n}}= \lim\limits_{n\to\infty}\frac{1}3\frac{(n+1)^3}{n^3} =\lim\limits_{n\to\infty}\frac{1}3\frac{n^3+3n^2+3n+1}{n^3}=\frac{1}3<1,$

we conclude that the series $\sum_{n=1}^{ \infty}\frac{n^3}{3^n}$ is covergent, and hence by Weiestrass M test the series $\sum_{n=1}^{ \infty} n^3 x^n$ converges uniformly on the interval $[-\frac{1}{3},\frac{1}{3}].$