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]

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]

but the second series is raised to the 2n+1 power, not the n power. This has to be addressed, too.

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]