Let's first find the Maclaurin series expansion for sinhxsinhx:
f(x)=sinhx=(e^x-e^-x)/2, f(0)=(e^0-e^0)/2=0f(x)=sinhx=ex−e−x2,f(0)=e0−e02=0
f'(x)=coshx=(e^x+e^-x)/2, f'(0)=(e^0+e^0)/2=1
f''(x)=sinhx, f''(0)=0
f'''(x)=coshx, f'''(0)=1
f^((4))(x)=sinhx, f^((4))(0)=0
f^((5))(x)=coshx, f^((5))(0)=1
So, we see a pretty consistent pattern of alternating zeroes and ones. Let's write out the first few terms of the series:
The Maclaurin Series expansion is given by
f(x)=sum_(n=0)^oof^((n))(0)x^n/(n!)=f(0)+f'(0)x+f''(0)x^2/(2!)+...
So, for our function, we get
sinhx=0+x+0x^2+x^3/(3!)+0x^4+x^5/(5!)+...
If we ignore the terms involving zero, we see
sinhx=x+x^3/(3!)+x^3/(3!)+...
So, we want odd exponents and odd factorials starting at 1, so the summation is
sinhx=sum_(n=0)^oox^(2n+1)/((2n+1)!)
To find the radius of convergence, we'll use the Ratio Test, where
a_n=x^(2n+1)/((2n+1)!)
lim_(n->oo)|a_(n+1)/a_n|=lim_(n->oo)|x^(2n+3)/((2n+3)!)*((2n+1)!)/x^(2n+1)|
We're going to want the factorials to cancel out. So, strip some terms out of the larger factorial:
(2n+3)! = (2n+3)(2n+1)(2n+1)!
So we have
lim_(n->oo)|(x^(2n+3)cancel((2n+1)!))/(x^(2n+1)(2n+3)(2n+2)cancel((2n+1)!))|
x^(2n+3)/x^(2n+1)=x^2
So,
|x^2|lim_(n->oo)1/((2n+3)(2n+2))<1 results in convergence.
The limit goes to 0. Thus, this quantity is always 0<1 regardless of what we pick for x. We have convergence for all real numbers, IE, R=oo