# What is the sum of the series 1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^n)/(n!))+...?

May 10, 2018

2

#### Explanation:

By definition of the exponential function:

e^Omega = sum_(k = 0)^oo Omega^k/(k!) = 1 + Omega + (Omega^2)/(2!) + ... + (Omega^n)/(n!) + ...

Look at the pattern:

• $\Omega = \ln 2$

And so you are evaluating:

• $\textcolor{red}{{e}^{\ln \left(2\right)} = 2}$