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