Question

Using wiestrass M test show that the following series converges uniformly Sigma infinity n=1 n^3 x^n x belongs to[-1/3,1/3]

Answer 1

The series

$\quad \quad \quad \quad \quad \quad \sum _{ n=1 }^{ \infty }{ { n }^{ 3 }{ x }^{ n }{x} } \quad \quad \quad \quad \quad \quad \quad \quad \quad \ (1)$

is a power series , the radius of convergence of which is 

$r=\ \lim _{ n\rightarrow \infty }{ \frac { { n }^{ 3 } }{ { (n+1) }^{ 3 } } } =1$

$\forall \ n\ge 1\quad and\quad \forall x\in \left[\frac{-1}{3},\frac { 1 }{ 3 } \right] :\ 0\le { n }^{ 3 }{ x }^{ n }{x}\le \frac { { n }^{ 3 } }{ { 3 }^{ n } } \ \quad \quad \quad \quad \ \quad (2)$

The series (1) converges in the interval $(-1,1)$. The point $x=\frac{1}{3}$ belongs to $(-1,1)$. Consequently series

$\quad\quad \quad\quad\quad\quad\quad\sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 3 }^{ n } } } \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad\quad\quad\quad\ (3)$

 

is convergent. According to the Weierstrass' M-Test , the series​ $\sum _{ n=1 }^{ \infty }{ { n }^{ 3 }{ x }^{ n }{x} }$

 converges uniformly in the interval $\left[ \frac{-1}{3},\frac { 1 }{ 3} \right]$ .

Option 2) The convergence of series (3) can be proved by the d'Alembert criterion 

${ a }_{ n }:=\frac { { n }^{ 2 } }{ { 3 }^{ n } } ,\lim _{ n\rightarrow \infty }{ \frac { { a }_{ n+1 } }{ { a }_{ n } } } =\lim _{ n\rightarrow \infty }{ \frac { { (n+1) }^{ 2 } }{ { 3 }^{ n+1 } } \cdot \frac { { 3 }^{ n } }{ { n }^{ 2 } } } =\frac { 1 }{ 3 } <1$

Given the fulfillment of (2), according to the Weierstrass' M-Test, we obtain the series (1) converges uniformly in the interval  $[\frac{-1}{3},\frac{1}{3}]$ .