Question

Use Weierstrass’ M-Test to prove that the series
Infinity
∑ n^2x^n
n=1
converges uniformly in
the interval .
[0,1/5]

Answer 1

Option 1) The series

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

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

$r=\ \lim _{ n\rightarrow \infty }{ \frac { { n }^{ 2 } }{ { (n+1) }^{ 2 } } } =1$

$\forall \ n\ge 1\quad and\quad \forall x\in \left[ 0,\frac { 1 }{ 5 } \right] :\ 0\le { n }^{ 2 }{ x }^{ n }\le \frac { { n }^{ 2 } }{ { 5 }^{ n } } \ \quad \quad \quad \quad \ \quad (2)$

The series (1) converges in the interval $(-1,1)$. The point $x=\frac{1}{5}$ belongs to $(-1,1)$. Consequently series

$\quad\quad \quad\quad\quad\quad\quad\sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 5 }^{ n } } } \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad\quad\quad\quad\ (3)$

 

is convergent. According to the Weierstrass' M-Test , the series​ $\sum _{ n=1 }^{ \infty }{ { n }^{ 2 }{ x }^{ n } }$

 converges uniformly in the interval $\left[ 0,\frac { 1 }{ 5 } \right]$ .

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

${ a }_{ n }:=\frac { { n }^{ 2 } }{ { 5 }^{ n } } ,\lim _{ n\rightarrow \infty }{ \frac { { a }_{ n+1 } }{ { a }_{ n } } } =\lim _{ n\rightarrow \infty }{ \frac { { (n+1) }^{ 2 } }{ { 5 }^{ n+1 } } \cdot \frac { { 5 }^{ n } }{ { n }^{ 2 } } } =\frac { 1 }{ 5 } <1$

Given the fulfillment of (2), according to the Weierstrass' M-Test, we obtain the series (1) converges uniformly in the interval  $[0,\frac{1}{5}]$ .