Question

Find maclaurin series

Answer 1

Recall the series expansion of the exponential function,

$\displaystyle e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

By comparison, the given sum is $e^x$ evaluated at x = ln(2), so

$\displaystyle 1 + \ln(2) + \frac{\ln^2(2)}{2!} + \cdots = \sum_{n=0}^\infty \frac{\ln^n(2)}{n!} = e^{\ln(2)} = \boxed{2}$