How do you find the taylor series series for ((e^x)-1)/x at c=0?

Apr 13, 2018

 (e^x-1)/x = 1 + x/(2!) + x^2/(3!) + x^3/(4!) + x^4/(5!) + ...

Explanation:

Whilst we could start from first principles and derive using the MacLaurin formula:

 f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...

This would seem to be to a significant amount of work when we can utilise the well known series for ${e}^{x}$

 e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...

From which we get:

 (e^x-1)/x = ({1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...}-1)/x

 \ \ \ \ \ \ \ \ \ \ \ = (x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ...)/x

 \ \ \ \ \ \ \ \ \ \ \ = 1 + x/(2!) + x^2/(3!) + x^3/(4!) + x^4/(5!) + ...