Question

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

Answer 1

Infinite.

Explanation:

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`.