What is the radius of convergence of $sum_1^oo (-1^n*x^(2n) )/ ((2n)!)$ ?

Question

3997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-01T11:21:19

Infinite.

This is the Maclaurin series for $cos(x) - 1$ , which converges for all Real values of $x$ .

Let $a_n = ((-1)^n*x^(2n))/((2n)!)$

Then:

$a_(n+1)/a_n = (((-1)^(n+1)*x^(2n+2))/((2n+2)!))/(((-1)^n*x^(2n))/((2n)!))$

$=-((2n)!)/((2n+2)!) x^2= -x^2/((2n+1)(2n+2))$

So for example, if $n > abs(x)$ then $abs(a_(n+1)/a_n) < 1/4$ .

So the tail of $sum_(n=1)^oo ((-1)^n*x^(2n))/((2n)!)$ will converge faster than a geometric series with common ratio $1/4$ .

What is the radius of convergence of sum_1^oo (-1^n*x^(2n) )/ ((2n)!)sum_1^oo (-1^n*x^(2n) )/ ((2n)!)?