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 :)