For the function f(x)=(e^x-1)/x, find f'(2) and use it to show that the sum of n/(n+1)! from n=1 to infinity equals 1.

To find the derivative of the function f(x) = (e^x - 1)/x, we can use the quotient rule. The quotient rule states that if we have a function u(x) = g(x)/h(x), then the derivative of u(x) is given by:

u'(x) = (g'(x)h(x) - g(x)h'(x))/[h(x)]^2.

Here, g(x) = e^x - 1 and h(x) = x. Let's calculate the derivatives:

g'(x) = d/dx (e^x - 1) = e^x,
h'(x) = d/dx (x) = 1.

Now, substituting these values into the quotient rule formula, we get:

f'(x) = [(e^x)(x) - (e^x - 1)(1)]/[x]^2
= [(e^x)x - (e^x - 1)]/x^2.

To find f'(2), we substitute x = 2 into the expression for f'(x):

f'(2) = [(e^2)(2) - (e^2 - 1)]/2^2
= [2e^2 - (e^2 - 1)]/4
= (e^2 + 1)/4.

Now, we want to find the sum of n/(n+1)! from n = 1 to infinity:

S = 1/2! + 2/3! + 3/4! + ...

We can rewrite each term as:

n/(n+1)! = n/[n!(n+1)]
= 1/n! - 1/[(n+1)!].

Now, let's rewrite the sum S using this new representation:

S = (1/1!) - (1/2!) + (1/2!) - (1/3!) + (1/3!) - (1/4!) + ...

Notice that most terms cancel each other out, leaving us with only the first term (1/1!), which simplifies to 1.

So, the sum of n/(n+1)! from n = 1 to infinity is equal to 1.