alejandrito20 Posted May 15, 2011 Share Posted May 15, 2011 In the serie [math] \sum_0^{\infty} a_n (x - c)^n [/math], the radius of convergency is: [math]R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|[/math]. My problem is : Find the radius of convergency when: [math] \sum_0^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot x^{2n+1} [/math] i don't understand mainly who is [math]a_n[/math]. The answer is [math]R= \infty[/math] Link to comment Share on other sites More sharing options...
DrRocket Posted May 16, 2011 Share Posted May 16, 2011 In the serie [math] \sum_0^{\infty} a_n (x - c)^n [/math], the radius of convergency is: [math]R= \lim_{n \to \infty } |\frac{a_n}{a_{n+1}}|[/math]. My problem is : Find the radius of convergency when: [math] \sum_0^{\infty} \frac{(-1)^n}{(2n+1)!} \cdot x^{2n+1} [/math] i don't understand mainly who is [math]a_n[/math]. The answer is [math]R= \infty[/math] [math] a_n= \frac{(-1)^n}{(2n+1)!} [/math] and [math]c=0[/math] Link to comment Share on other sites More sharing options...
Bignose Posted May 16, 2011 Share Posted May 16, 2011 but the second series is raised to the 2n+1 power, not the n power. This has to be addressed, too. Link to comment Share on other sites More sharing options...
DrRocket Posted May 16, 2011 Share Posted May 16, 2011 but the second series is raised to the 2n+1 power, not the n power. This has to be addressed, too. Go back to the proof using the ratio test. You will find convergence when x^2 is less than the limit indicated in the OP which, since the limit is infinite means when |x| is arbitrary. You can also look at the associated series in terms of complex numbers (note that the series is sin (x)). [math] sin(x) = Im ( \sum_{n=0}^ \infty \dfrac {(ix)^n}{n!})[/math] This all boils down to the fact that the power series for the exponential function has an infinitevradius of convergence. [math]exp(z) = \sum_{n=0}^ \infty \dfrac {z^n}{n!}[/math] Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now