# series convergence and radius of convergence

• Nov 14th 2008, 07:18 PM
drpawel
series convergence and radius of convergence
Can you help mr to solve that
let c0,c1,c2...in R prove that if lim when n goes to infinity of abs(c(subn)/c(sub n+1)) exists,it is equal to the radius of convergence of the power series c(subn)x^n
• Nov 15th 2008, 08:07 AM
Mathstud28
Quote:

Originally Posted by drpawel
Can you help mr to solve that
let c0,c1,c2...in R prove that if lim when n goes to infinity of abs(c(subn)/c(sub n+1)) exists,it is equal to the radius of convergence of the power series c(subn)x^n

Ok let $\displaystyle \left\{c_n\right\}$ be a sequence in $\displaystyle \mathbb{R}$. Ok let us split this up into three cases: $\displaystyle \lim_{n\to\infty}\frac{c_{n}}{c_{n+1}}=L\ne{0}$, $\displaystyle \lim_{n\to\infty}\frac{c_{n}}{c_{n+1}}=\infty$, and $\displaystyle \lim_{n\to\infty}\frac{c_{n}}{c_{n+1}}=0$.

Case 1

Suppose we have a series of the form $\displaystyle \sum{c_nx^n}$. Then to determine the radius of convergence I apply the Ratio test. So $\displaystyle \lim_{n\to\infty}\left|\frac{c_{n+1}x^{n+1}}{c_{n} x^n}\right|=|x|\lim_{n\to\infty}\frac{c_{n+1}}{c_n }<1$. So now by our assumption above this is equivalent to $\displaystyle |x|\frac{1}{L}<1\implies|x|<L$, thus the radius of convergence of case 1 is $\displaystyle L$ and we are done.

Case 2
Now suppose we have the same series but we know have that $\displaystyle \lim_{n\to\infty}|x|\frac{c_{n}}{c_{n+1}}\implies| x|\cdot{0}<1$, which is an identity for $\displaystyle |x|<\infty$. Thus the radius of convergence is $\displaystyle \infty$.

Case 3
Now once again we reach that $\displaystyle |x|\lim_{n\to\infty}\frac{c_n}{c_{n+1}}=|x|\cdot{\ infty}<1$ which only occurs when $\displaystyle x=0$.

The last two I did more holistically than the first, if you need more rigor I leave that up to you.