Answer to Question #210797 in Real Analysis for Nikhil

Question

Show that the series Σ x/(1+n^2.x^2) is uniformly convergent using Weierstrass m test

Answer 1

Since for each $n\in N$ and $x\in \left[ 1,\alpha \right]$ , we have $1+{ n }^{ 2 }{ x }^{ 2 }\ge { n }^{ 2 }\ ,\ x\le \alpha$ ,then

$0\le \frac { x }{ 1+{ n }^{ 2 }{ x }^{ 2 } } \le \frac { \alpha }{ { n }^{ 2 } }$

 .

Setting ${ f }_{ n }(x)=\frac { x }{ 1+{ n }^{ 2 }{ x }^{ 2 } } ,\ { M }_{ n }=\frac { \alpha }{ { n }^{ 2 } }$ we have for each $n\in N$ and $x\in \left[ 1,\alpha \right]$ : 

$\left| { f }_{ n }(x) \right| ={ f }_{ n }(x)\le { M }_{ n }$ . The series $\sum _{ n=1 }^{ \infty }{ { M }_{ n }\ } =\alpha \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }$ converges, because the series 

$\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } }$ is p-series (p=2). By the Weierstrass M-Test , the series

$\sum _{ n=1 }^{ \infty }{ { f }_{ n }(x)= } \sum _{ n=1 }^{ \infty }{ \frac { x }{ 1+{ n }^{ 2 }{ x }^{ 2 } } }$

converges uniformly in $\left[ 1,\alpha \right]$ .