f(x)=e^(-2x) Prove that the Maclaurin series for the function converges to the function for all x

Maclaurin series is a special case of Taylor series that is centered at c=0 . The expansion of the function about 0 follows the formula:
f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n
 or
f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0))/(3!)x^3+(f^4(0))/(4!)x^4 +...
To determine the Maclaurin polynomial of degree n=5 for the given function f(x)=e^(-2x) , we may apply the formula for Maclaurin series..
To list derivative functions f^n(x) , we may apply derivative formula for exponential function: d/(dx) e^u = e^u * (du)/(dx) .
f(x)=e^(-2x)
Let u =-2x then (du)/(dx)= -2 .
d/(dx) e^(-2x) = e^(-2x) *(-2)
                 = -2e^(-2x)
Applying d/(dx) e^(-2x)= -2e^(-2x)   for each derivative function, we get:
f'(x) = d/(dx) e^(-2x)
        =-2e^(-2x)
f^2(x) = d/(dx) (- 2e^(-2x))
            =-2 d/(dx) (e^(-2x))
            =(-2)*(-2e^(-2x))
            =4e^(-2x)
f^3(x) = d/(dx) (4e^(-2x))
            =4d/(dx) (e^(-2x))
            =4*(-2e^(-2x))
             =-8e^(-2x)
 f^4(x) = d/(dx) (- 8e^(-2x))
             =-8 d/(dx) (e^(-2x))
             =(-8)*(-2e^(-2x))
             =16e^(-2x)
 Plug-in x=0 for each f^n(x) , we get:
 f(0) =e^(-2*0) =1
 f'(0) =-2e^(-2*0)=-2
 f^2(0) =4e^(-2*0)=4
 f^3(0) =-8e^(-2*0)=-8
 f^4(0) =16e^(-2*0)=16
Note: e^(-2*0) = e^0 = 1.
 Plug-in the values on the formula for Maclaurin series, we get:
 sum_(n=0)^oo (f^n(0))/(n!) x^n
 = 1+(-2)/(1!)x+4/(2!)x^2+(-8)/(3!)x^3+16/(4!)x^4+...
 = 1-2/(1!)x+4/(2!)x^2-8/(3!)x^3+16/(4!)x^4+...
  =sum_(n=0)^oo (-2)^n/(n!)x^n
 =sum_(n=0)^oo (-2x)^n/(n!)
To determine the interval of convergence for the Maclaurin series: sum_(n=0)^oo (-2x)^n/(n!) , we may apply Ratio Test.  
In Ratio test, we determine the limit as: lim_(n-gtoo)|a_(n+1)/a_n| = L.
The series converges absolutely when it satisfies Llt1 .
For the  Maclaurin series: sum_(n=0)^oo (-2x)^n/(n!) , we have:
a_n=(-2x)^n/(n!)
Then,
1/a_n= (n!)/(-2x)^n
a_(n+1)=(-2x)^(n+1)/((n+1)!)
            =((-2x)^n*(-2x)^1)/((n+1)*(n!))
              =((-2x)^n(-2x))/((n+1)*(n!))
Applying the Ratio test, we set-up the limit as:
lim_(n-gtoo)|a_(n+1)/a_n|=lim_(n-gtoo)|a_(n+1)*1/a_n|
                          =lim_(n-gtoo)|((-2x)^n(-2x))/((n+1)*(n!))*(n!)/(-2x)^n|
Cancel out common factors: (-2x)^n  and (n!) .
lim_(n-gtoo)|(-2x)/(n+1)|
Evaluate the limit.
lim_(n-gtoo)|(-2x)/(n+1)|=|-2x| lim_(n-gtoo)|1/(n+1)|
                        =|2x|lim_(n-gtoo)1/(n+1)
                        =|2x|* 1/oo
                        = |2x|*0
                        =0
The L=0 satisfies  Llt1 for all x .
Thus, the Maclaurin series: sum_(n=0)^oo (-2x)^n/(n!) is absolutely converges for all x .
Interval of convergence: -ooltxltoo