f(x)=cosx Prove that the Maclaurin series for the function converges to the function for all x

Maclaurin series is a special case of Taylor series which is centered at c=0 . We follow the formula:
f(x)=sum_(n=0)^oo (f^n(0))/(n!)x^n
or
f(x) = f(0)+ f'(0)x +(f^2(0))/(2!)x^2 +(f^3(0))/(3!)x^3 +(f^4(0))/(4!)x^4 +...
To list the f^n(x) , we may apply derivative formula for trigonometric functions: 
d/(dx) sin(x) = cos(x) and d/(dx)cos(x) = -sin(x).
f(x)=cos(x)
f'(x)=d/(dx)cos(x) = -sin(x)
f^2(x)=d/(dx) -sin(x) = -cos(x)
f^3(x)=d/(dx) -cos(x)= - (-sin(x))= sin(x)
f^4(x)=d/(dx)d/(dx) sin(x) = cos(x)
Plug-in  x=0 , we get:
f(0)=cos(0) =1
f'(0) = -sin(0)=0
f^2(0) = -cos(0)=-1
f^3(0)=sin(0)=0
f^4(0)= cos(0) =1
Note: cos(0)= 1 and sin(0)=0 .
Plug-in the f^n(0) values on the formula for Maclaurin series, we get:
cos(x) =sum_(n=0)^oo (f^n(0))/(n!)x^n
              =1 +0*x+(-1)/(2!)x^2+(0)/(3!)x^3+(1)/(4!)x^4+...
               =1 +0-1/2x^2+0/6x^3 +1/24x^4+...
              =1 +0-1/2x^2+0 +1/24x^4+...
               =1 -1/2x^2 +1/24x^4+...
                = sum_(n=0)^oo ((-1)^n x^(2n))/((2n)!)
To determine the interval of convergence, we apply Ratio test.
In ratio test, we determine a limit as lim_(n-gtoo)| a_(n+1)/a_n| =L where a_n!=0 for all ngt=N .
The series sum a_n is a convergent series when L lt1 .
From the Maclaurin series of cos(x) as sum_(n=0)^oo ((-1)^n x^(2n))/((2n)!) , we have:
a_n= ((-1)^n x^(2n))/((2n)!) then 1/a_n=((2n)!) /((-1)^n x^(2n))
Then, a_(n+1) =(-1)^(n+1) x^(2(n+1))/((2(n+1))!)
                      =(-1)^(n+1) x^(2n+2)/((2n+2)!)
                      =(-1)^n*(-1)^1 (x^(2n)*x^2)/((2n+2)(2n+1)(2n)!)
                      = ((-1)^n*(-1)x^(2n)*x^2)/((2n+2)(2n+1)(2n)!)
We set up the limit lim_(n-gtoo)| a_(n+1)/a_n| as:
lim_(n-gtoo) |a_(n+1)/a_n| =lim_(n-gtoo) |a_(n+1) * 1/a_n |
=lim_(n-gtoo) |((-1)^n*(-1)^1* x^(2n)*x^2)/((2n+2)(2n+1)(2n)!)*((2n)!) /((-1)^n x^(2n))|
Cancel out common factors: (-1)^n, (2n)!, and x^(2n) , the limit becomes;
lim_(n-gtoo) |(-x^2)/((2n+2)(2n+1))|
Evaluate the limit.
lim_(n-gtoo) |- x^2/((2n+2)(2n+1))|=|-x^2/2| lim_(n-gtoo) 1/((2n+2)(2n+1))
                                           =|-x^2/2|*1/ oo
                                           =|-x^2/2|*0
                                            =0
The L=0 satisfy the Llt1 for every x .
Therefore, Maclaurin series of cos(x) as sum_(n=0)^oo (-1)^n x^(2n)/((2n)!) converges for all x.
Interval of convergence: -ooltxltoo .