A calculus problem by Muhammad Shariq

We first look at the Taylor Series and observe that:

$\large \displaystyle \sum_{n=0}^\infty \frac{x^{n+1}}{n!}=x+x^2+\frac{x^3}{2!}+...+\frac{x^{n+1}}{n!}=x(1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!})=xe^x$

Then:

$\large f(x)=\ln(xe^x)=\ln(x)+\ln(e^x)=\ln(x)+x \implies f'(x)=\frac{1}{x}+1 \implies f'(1)=\frac{2}{1}$

Therefore $\large a+b=2+1=\boxed{3}$ .