0
$\begingroup$

Lets say that for every $n\in \mathbb{N}$$,a_n\ne0$.

If, $$\lim_{n\to 0}\frac{a_{n+1}}{a_n}=l$$ how do I prove that the radius of the convergence of the series:$$\sum_{n=0}^{\infty}a_nx^n$$ is $\frac{1}{|l|}$ if $l\ne 0$ and $\infty$ is $l=0$.

I have no idea how to prove it...

Thank you!

$\endgroup$
1
$\begingroup$

Hint : Apply ratio test to $$\sum_{n=1}^{\infty} a_n x^n$$ .

EDIT : No problem if $n=0$ or $n=1$ as finite terms doesn't affect the convergence.

$\endgroup$
  • $\begingroup$ It's matter if it's from $n=0$ and not $n=1$? $\endgroup$ – CS1 Dec 1 '13 at 12:46
  • 1
    $\begingroup$ @YoavFridman: convergence of series only depends on the tail of the series, not on the beginning. $\endgroup$ – robjohn Dec 1 '13 at 12:47
  • $\begingroup$ I don't know if $l>1$ or $l<1$... $\endgroup$ – CS1 Dec 1 '13 at 12:47
  • 1
    $\begingroup$ You must determine which $x$'s which make the ratio less than $1$ , and hence have your radius of convergence $\endgroup$ – Mahmoud Abdulrazek Dec 1 '13 at 13:10
1
$\begingroup$

Let $R=\frac {1}{|l|}$.For example if $$|x|<R=>|a_nx^n|=|a_n||x|^n=|a_n|R^n\frac {|x|^n}{R^n}\leq M (\frac {|x|}{R})^n$$ and $\sum (\frac {|x|}{R})^n$ is a geometric series.

Try the rest.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.