series convergence and radius of convergence

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

Case 1

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

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

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

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