How do you find the sum of the series 1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^2)/(n!))+...?

The sum is $2$.
Let's write the McLaurin Series of $y = {2}^{x}$:
2^x=2^0+(2^0ln2)/(1!)x+(2^0(ln2)^2)/(2!)x^2+...+(2^0(ln2)^n)/(n!)x^n+....
If we put in this series $x = 1$, on the left the result is $2$, on the right the result is the series of the exercise.