# What is the taylor series of xe^x?

Apr 26, 2017

 xe^x = x + x^2 + x^3/(2!)+x^4/(3!) + x^5/(4!) + ...
 \ \ \ \ \ = sum_(n=0)^oo x^(n+1)/(n!)

#### Explanation:

We can start with the well known Maclaurin series for ${e}^{x}$

 e^x = 1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ...
 \ \ \ = sum_(n=0)^oo x^n/(n!)

So then multiplying by $x$ we have:

 xe^x = x{1 + x + x^2/(2!)+x^3/(3!) + x^4/(4!) + ... }
 \ \ \ \ \ = x + x^2 + x^3/(2!)+x^4/(3!) + x^5/(4!) + ...
 \ \ \ \ \ = sum_(n=0)^oo x*x^n/(n!)
 \ \ \ \ \ = sum_(n=0)^oo x^(n+1)/(n!)