f(x)=coshx 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 +(f^5(0))/(5!)x^5+...
To determine the Maclaurin series for the given function f(x)=cosh(x) , we may apply the formula for Maclaurin series.
To list f^n(x),  we may follow the derivative formula for hyperbolic trigonometric functions:  d/(dx) cosh(x) = sinh(x) and 
d/(dx) sinh(x) = cosh(x).
f(x) =cosh(x)
f'(x) = d/(dx) cosh(x)= sinh(x)
f^2(x) = d/(dx) sinh(x)= cosh(x)
f^3(x) = d/(dx) cosh(x)= sinh(x)
f^4(x) = d/(dx) sinh(x)=cosh(x)
f^5(x) = d/(dx) cosh(x)=sinh(x)
f^6(x) = d/(dx) sinh(x)=cosh(x)
Note: When n= even then f^n(x)=cosh(x).
When n= odd then f^n(x)=sinh(x).
Plug-in x=0 on each f^n(x) , we get:
f'(0) =cosh(0)=1
f'(0) =sinh(0)=0
f^2(0) =cosh(0)=1
f^3(0) =sinh(0)=0
f^4(0) =cosh(0)=1
 f^5(0) =sinh(0)=0
 f^6(0) =cosh(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+0/(1!)x+1/(2!)x^2+0/(3!)x^3+1/(4!)x^4 +0/(5!)x^5+1/(6!)x^6+ ...
=1+0+1/(2!)x^2+0+1/(4!)x^4+0+1/(6!)x^6+ ...
=1+/(2!)x^2+1/(4!)x^4+1/(6!)x^6+ ...
=sum_(n=0)^oo x^(2n)/((2n)!)
The Maclaurin series is sum_(n=0)^oo x^(2n)/((2n)!) for the function f(x)=cosh(x) .
To determine the interval of convergence for the Maclaurin series: sum_(n=0)^oo x^(2n)/((2n)!) , 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 .
In the Maclaurin series: sum_(n=0)^oo x^(2n)/((2n)!) , we have:
a_n=x^(2n)/((2n)!)
Then,
1/a_n=((2n)!)/x^(2n)
a_(n+1)=x^(2(n+1))/(2(n+1)!)
           =x^(2n+2)/((2n+2)!)
           =(x^(2n)*x^2)/((2n+2)(2n+1)((2n)!))
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)|(x^(2n)*x^2)/((2n+2)(2n+1)((2n)!))*((2n)!)/x^(2n)|
Cancel out common factors: x^(2n) and (2n)! .
lim_(n-gtoo)|x^2/((2n+2)(2n+1))|
Evaluate the limit.
lim_(n-gtoo)|x^2/((2n+2)(2n+1))| = |x^2|lim_(n-gtoo)|1/((2n+2)(2n+1))|
                                 =|x^2|*1/oo
                                = |x^2|*0
                                =0
The L=0 satisfies Llt1 for all x . Thus, the Maclaurin series: sum_(n=0)^oo x^(2n)/((2n)!) is absolutely converges for all x .
Interval of convergence: -ooltxltoo .