Answer to Question #320287 in Real Analysis for Vikram

Question

Using weiestrass M-test, show that the following series converges uniformly.

∞

∑n^3X^n,X belongs to[-1/3,1/3]

n=1

Answer 1

ANSWER

Denote $f_{n}(x) =n^{3}x^{n}.$ If $\ x\in\left [ -\frac{1}{3} ,\frac{1}{3}\right ]$ , then

$\left |f _{n}(x) \right |\leq\frac{n^{3}}{3^{n}}$ .

Let $a_{n}=\frac{n^{3}} {3^{n}}$ .

$\frac{a_{n+1}}{a_{n}}=\frac{(n+1) ^{3}\cdot3^{n}}{3^{n+1}\cdot n^{3}}= \frac{1}{3}\cdot \left ( 1+\frac{1}{n} \right )^{3}$ .

$\lim _{n\rightarrow\infty}\frac{1}{3}\cdot \left ( 1+\frac{1}{n} \right )^{3}=$ $\frac{1}{3}\cdot$ $\lim _{n\rightarrow\infty}\ \left ( 1+\frac{1}{n} \right )^{3}=$ $\frac{1}{3 }\cdot \left ( \lim_{n\rightarrow\infty}\left ( 1+\frac{1}{n} \right ) \right )^{3}=\frac{1}{3}<1$

Because   $\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=\frac{1}{3}<1\$ , then the series $\sum_{n=1}^{\infty}a_{n}$ converges (by the Ratio Test).

Thus, using the Weierstrass M-Test , we make sure that the series $\sum_{n=1}^{\infty}n^{3}x^{n}$ converges uniformly on $\left [ -\frac{1}{3} ,\frac{1}{3}\right ]$