Let's first find the Maclaurin series expansion for #sinhx#:
#f(x)=sinhx=(e^x-e^-x)/2, f(0)=(e^0-e^0)/2=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#
