Infinite series (Analysis)

• Dec 7th 2008, 03:26 PM
thahachaina
Infinite series (Analysis)
Hey guys,

How do I prove that (1/nx) --> 0 pointwise but not uniformly on (0,1) as n tends to infinity.

• Dec 7th 2008, 08:57 PM
CaptainBlack
Quote:

Originally Posted by thahachaina
Hey guys,

How do I prove that (1/nx) --> 0 pointwise but not uniformly on (0,1) as n tends to infinity.

On $(0,1)$ and any $\varepsilon>0$

$
\frac{1}{nx}<\varepsilon,\ \ \forall n>N=\left \lceil \frac{1}{\varepsilon x}\right \rceil
$

which proves pointwise convergence.

For uniform convergence for $\varepsilon>0$ we need to find an $N$ such that for all $x\in (0,1):$

$
\frac{1}{nx}<\varepsilon,\ \ \forall n>N$

But for $x<\frac{1}{n\varepsilon}$ where $n>N;\ \frac{1}{nx}>\varepsilon$ which contradicts what would have to be true for uniform convergence to hold.

CB
• Nov 14th 2012, 12:13 PM
dave0147
Re: Infinite series (Analysis)
I'm lost again! What is < br >?
• Nov 17th 2012, 02:09 AM
hollywood
Re: Infinite series (Analysis)
I think that's just a formatting code that's no longer used - nothing mathematical about it. Here's CB's reply without the "br" formatting codes:

-------------------------

On $(0,1)$ and any $\varepsilon>0$,

$\frac{1}{nx}<\varepsilon,\ \ \forall n>N=\left \lceil \frac{1}{\varepsilon x}\right \rceil$

which proves pointwise convergence.

For uniform convergence for $\varepsilon>0$ we need to find an N such that for all $x\in (0,1)$:

$\frac{1}{nx}<\varepsilon,\ \ \forall n>N$

But for $x<\frac{1}{n\varepsilon}$ where $n>N;\ \frac{1}{nx}>\varepsilon$ which contradicts what would have to be true for uniform convergence to hold.

CB

-------------------------

By the way, why did you resurrect this 4-year-old thread?

- Hollywood