The terms of the infinite series sum: ∑f(n)(c)(x-c)n/n!, where c is a constant value to base the series on and integer n is such that 0≤n<∞. When n=0 the term is simply f(c). The next term is df/dx|c(x-c)/1!. f(n)(c) is the nth derivative of f where f(x) in this case is sinh(x)/x.
We want a series for sinh(x)/x. sinh(x)=½(ex-e-x). When x=0 sinh(x)=0, and sinh(x)/x is undefined, therefore we need to find c≠0 such that sinh(c)/c is defined.
But there is another way. We already have a series for ex=1+x+x2/2!+...+xn/n!+... which can be derived if f(x)=ex and c=0, because all derivatives of f(x) are ex. This is one of the simplest Taylor or MacLaurin series.
Therefore e-x=1-x+x2/2!-...+(-1)nxn/n!+...
From this we have a series for sinh(x):
½{(1+x+x2/2!+...+xn/n!+...)-(1-x+x2/2!-...+(-1)nxn/n!+...)}=
½(2x+2x3/3!+2x5/5!+...2xn/n!+...)=x+x3/3!+x5/5!+...xn/n!+...
To find a series for sinh(x)/x we simply divide by x:
1+x2/3!+x4/5!+...xn-1/n!+... where integer n>0.
Interestingly, plugging in x=0 gives f(0)=1, so sinh(x)/x=1 in the limit as x→0.