Since we have a factorial in the denominator, we use the ratio test , since this makes the simplifications a bit easier. This formula is:
#lim_(n->oo)(a_(n+1)/a_n)#
If this is < 1, your series converges
If this is > 1, your series diverges
If this is = 1, your test is inconclusive
So, let's do this:
#lim_(k->oo)abs((-1)^(k+1)(x^(2k+2)/((2k+2)!))*(-1)^k((2k)!)/(x^(2k))#
Note: Be very careful about how you plug in your (k+1). 2k will turn into 2(k+1), NOT 2k+1.
I multiplied by the reciprocal of #x^(2k)/((2k)!)# instead of dividing just to make the work a bit easier.
Now, let's algebra. Due to the absolute value, our alternating terms (i.e. #(-1)^k#) are just going to cancel out, since we will always have a positive answer:
#=> lim_(k->oo)abs((x^(2k+2)/((2k+2)!))*((2k)!)/(x^(2k))#
We can cancel our #x^(2k)#'s:
#=> lim_(k->oo)abs((x^2/((2k+2)!))*((2k)!)#
Now we need to cancel out factorials.
Recall that #(2k)! = (2k) * (2k-1) * (2k-2) * (2k-3) * ... * 3 * 2 * 1#
Also, #(2k+2)! = (2k + 2) * (2k + 1) * (2k) * (2k - 1) * .... * 3 * 2 * 1#
Notice:
#(2k)! = color(red)((2k) * (2k-1) * (2k-2) * (2k-3) * ... * 3 * 2 * 1)#
#(2k+2)! = (2k + 2) * (2k + 1) * color(red)((2k) * (2k - 1) * .... * 3 * 2 * 1)#
As you can see, we #(2k)!# is essentially a part of #(2k + 2)!#. We can use this to cancel out every common term:
#((2k)!)/((2k+2)!) = cancel(color(red)((2k) * (2k-1) * (2k-2) * (2k-3) * ... * 3 * 2 * 1))/( (2k + 2) * (2k + 1) * cancel(color(red)((2k) * (2k - 1) * .... * 3 * 2 * 1))#
#= 1/((2k + 2)(2k + 1))#
This leaves
#=> lim_(k->oo)abs((x^2/((2k + 2) (2k + 1) ))#
Now, we can evaluate this limit. Note that since we're not taking this limit with respect to #x#, we can factor it out:
#=> abs(x^2 lim_(k->oo)(1/((2k + 2)(2k+1))) #
#=> abs(x^2 * 0) = 0#
So as you can see, this limit = 0, which is less than 1. Now, we ask ourselves: is there any value of #x# for which this limit would be ≥ 1? And the answer is no, since anything multiplied by 0 is 0.
So, since #lim_(k->oo)abs((x^(2k+2)/((2k+2)!))*((2k)!)/(x^(2k))) < 1# for all values of #x#, we can say that it has an interval of convergence of #(-oo, oo)#.
Hope that helped :)
