If there were a countable sequence of functions fn(x), eg f1(x), f2 (x), .......

what is the infimum limit (lim inf) of the set of funtions, is it the pointwise convergence function, ie lim inf for every value of x?

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- Jan 12th 2009, 12:43 PMdanirougeVery strange? Limit inferiror of a sequence of functions.
If there were a countable sequence of functions fn(x), eg f1(x), f2 (x), .......

what is the infimum limit (lim inf) of the set of funtions, is it the pointwise convergence function, ie lim inf for every value of x? - Jan 12th 2009, 02:16 PMlukaszh
Take an example:

$\displaystyle \left\{\frac{(x-2)^n}{5^n(3n+2)}\right\}_{n=1}^{\infty}$

That is a sequence:

$\displaystyle \left\{a_n(x-a)^n\right\}_{n=1}^{\infty}$

I can find a limit superior of a_n:

$\displaystyle \limsup_{n\to\infty}\frac{1}{5^n(3n+2)}=0$

and also limit inferior:

$\displaystyle \liminf_{n\to\infty}\frac{1}{5^n(3n+2)}=0$

This sequence is convergent because

$\displaystyle \limsup_{n\to\infty}a_n=\liminf_{n\to\infty}a_n=\l im_{n\to\infty}a_n$

The sequence

$\displaystyle \left\{\frac{(x-2)^n}{5^n(3n+2)}\right\}_{n=1}^{\infty}$

is convergent only for some x. You have to find the radius of convergence by the limit:

$\displaystyle r:=\limsup_{n\to\infty}\frac{1}{\sqrt[n]{|a_n|}}$

if $\displaystyle r\in\mathbb{R}$, then radius R is:

$\displaystyle R=\frac{1}{r}$ - Jan 13th 2009, 12:17 AMOpalg