# Question 35a7e

Feb 12, 2018

As mentioned in the comments below, this is the MacLaurin series for $f \left(x\right) = \cos \left(x\right)$, and we know that this converges on $\left(- \infty , \infty\right)$. However, if you wanted to see the process:

#### Explanation:

Since we have a factorial in the denominator, we use the ratio test , since this makes the simplifications a bit easier. This formula is:

${\lim}_{n \to \infty} \left({a}_{n + 1} / {a}_{n}\right)$

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. ${\left(- 1\right)}^{k}$) 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 ${x}^{2 k}$'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))

$= \frac{1}{\left(2 k + 2\right) \left(2 k + 1\right)}$

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))) 

$\implies \left\mid {x}^{2} \cdot 0 \right\mid = 0$

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 $\left(- \infty , \infty\right)$.

Hope that helped :)