Sumtegrals!

The expansion of the exponential function around zero is given by $e^x = \sum_{n=0}^\infty \dfrac{x^n}{n!}.$ Then, multiplying both sides by $x^2$ , we find that $x^2 e^x = \sum_{n=0}^\infty \dfrac{x^{n+2}}{n!}.$ After that, we integrate both sides with respect to $x$ in the interval $[0,2]$ . Note that we can interchange the integral with the sum, because the sum is absolutely convergent. $\int_0^2 x^2 e^x dx = \sum_{n=0}^\infty \left( \int_0^2 \dfrac{x^{n+2}}{n!} dx \right) = \sum_{n=0}^\infty \dfrac{2^{n+3}}{(n+3)n!}.$ Finally, we can see that our desired sum is given by $\sum_{n=0}^\infty \dfrac{2^{n}}{(n+3)n!} = \dfrac{1}{8} \int_0^2 x^2 e^x dx = \dfrac{1}{8} \left[ (x^2-2x+2) e^x \right]_{x=0}^2 = \boxed{ \dfrac{1}{4} (e^2 - 1) }.$

$\large\sum_{n=0}^\infty\frac{2^n}{(n+3)n!}=\large\sum_{n=0}^\infty\frac{2^n(n+2)(n+1)}{(n+3)!}=\large\sum_{n=3}^\infty\frac{2^{n-3}(n-1)(n-2)}{n!}=\frac{1}{8}\large\sum_{n=3}^\infty\frac{2^{n}(n-1)(n-2)}{n!}$

Consider the following infinite series for $e^x$ obtained by getting its zeroth, first, and second derivatives evaluated at $x=2$ . $e^2=\large\sum_{n=0}^\infty\frac{2^n}{n!}$ $e^2=\large\sum_{n=0}^\infty\frac{n(2)^{n-1}}{n!}$ $e^2=\large\sum_{n=0}^\infty\frac{n(n-1)(2)^{n-2}}{n!}$ The second equation is equivalent to $2e^2=\large\sum_{n=0}^\infty\frac{n(2)^{n}}{n!}$ and the third equation to $4e^2=\large\sum_{n=0}^\infty\frac{n(n-1)(2)^{n}}{n!}$

$\frac{1}{8}\large\sum_{n=3}^\infty\frac{2^{n}(n-1)(n-2)}{n!}=\frac{1}{8}(\large\sum_{n=0}^\infty\frac{n(n-1)(2)^{n}}{n!}-2\large\sum_{n=0}^\infty\frac{n(2)^{n}}{n!}+2\large\sum_{n=0}^\infty\frac{2^n}{n!}-2)=\frac{1}{8}(4e^{2}-2(2e^2)+2e^2-2)=\frac{1}{4}(e^2-1)$ Thus, $a+b=6$