What is the taylor series of #xe^x#?

1 Answer
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!) #