Question

Find maclaurin series

Answer 1

Recall the series expansion of the exponential function,

`\displaystyle e^x = \sum_(n=0)^\infty (x^n)/(n!)`

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

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