Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 55

$2$

Since, we know that the Maclaurin Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Thus, we have the form of the given series as an $e^x$ series. Here, in the given problem $x=\ln 2$ Thus, the sum of the series is: $e^{\ln 2}=2$