im learning about radius of convergence now but i still have yet to grasp its concept.

let "R hat" = (lim sup la_nl^1/n )^-1
assume that a_n is not 0 and the seq of positive numbers la_nl/la_n+1l converges to a limit. how do i prove that the lim la_nl/la_n+1l = 'R hat" where R hat is the radius of convergence of the series( sum from n = 1 to infinitiy a_n x^n?

thanks!

2. Originally Posted by alexandrabel90
im learning about radius of convergence now but i still have yet to grasp its concept.

let "R hat" = (lim sup la_nl^1/n )^-1
assume that a_n is not 0 and the seq of positive numbers la_nl/la_n+1l converges to a limit. how do i prove that the lim la_nl/la_n+1l = 'R hat" where R hat is the radius of convergence of the series( sum from n = 1 to infinitiy a_n x^n?

thanks!

u have Cauchy - Hadamard Th. saying that :

let the $R$ be radius of convergence of the power series $\displaystyle \sum _{n=1} ^{\infty} a_n x^n$, then it is worth ::

$\displaystyle R = \frac {1}{\lim_{n\to \infty} \sqrt[n] {|a_n|}}$

then if , $\displaystyle \lim_{n\to \infty} \sqrt[n] {|a_n|}=+\infty$ then radius is $R=0$ which means that power series are converging just in point $x=0$ (all power series are converging at least at point x=0)

or if $\displaystyle \lim_{n\to \infty} \sqrt[n] {|a_n|}=0$ then radius is $R=+\infty$ which means that series converge for all $x\in \mathbb{R}$.

P.S. you can depending on the situations use another formula for calculating radius of convergence (less work sometimes)

$\displaystyle R = \lim_{n\to \infty} |\frac {a_n}{a_{n+1}}|$

now if i understand your question ... you are referring to the Th and need the proof of it which says :

Th::

let the power series $\displaystyle \sum _{n=1} ^{\infty} a_n x^n$ is converging for some value $x_0 \neq 0$. Than series are absolutely convergent for all $x$ for which is $|x|< |x_0|$ and that series uniformly convergent for all $x$ for which $|x|\leq r$ where we have that $r<|x_0|$.

proof ::

let's assume that for $x_0\neq 0$ series $\displaystyle \sum _{n=1} ^{\infty} a_n x^n$ converge. That means that common member of the series must strive for zero, and thus he as a convergent sequence is limited. So there is constant $M>0$ such that for all $n\in \mathbb{N}$ this is true $|a_n x_0 ^n | \leq M$

now let the $|x| < |x_0|$ than is true that

$\displaystyle |a_n x ^n | = |a_n x_0 ^n \frac {x^n}{x_0 ^n} |= |a_n x_0 ^n | \cdot |\frac {x^n}{x_0 ^n} | \leq M |\frac {x^n}{x_0 ^n} |$

because $|\frac {x^n}{x_0 ^n} | <1$ than , series $\displaystyle \sum _{n=1} ^{\infty} |\frac {x^n}{x_0 ^n} |^n$ as the sum of the geometrical series is converging, which now means that and series $\displaystyle \sum _{n=1} ^{\infty} |a_n x^n |$ is converging, and that $\displaystyle \sum _{n=1} ^{\infty} a_n x^n$ is absolutely converge.

now we show second part

let $|x|\leq r$ where $r< |x_0|$ Based on the proof above we know that series $\displaystyle \sum _{n=1} ^{\infty} a_n r^n$ absolutely converge. As for the selected $x$ is valid ::

$|a_n x^n | = |a_n||x^n|\leq |a_n| r^n$

using Weierstrass criteria we have uniform convergence of the series $\displaystyle \sum _{n=1} ^{\infty} a_n x^n$

3. Originally Posted by alexandrabel90
im learning about radius of convergence now but i still have yet to grasp its concept.

let "R hat" = (lim sup la_nl^1/n )^-1
assume that a_n is not 0 and the seq of positive numbers la_nl/la_n+1l converges to a limit. how do i prove that the lim la_nl/la_n+1l = 'R hat" where R hat is the radius of convergence of the series( sum from n = 1 to infinitiy a_n x^n?

thanks!

$\frac{a_2}{a_1}\frac{a_3}{a_2}\frac{a_4}{a_3}...\f rac{a_{n+1}}{a_n}$