Need help in series summation: $\sum_{a=2}^{\infty}\frac{2^{a}}{a\cdot a!}$

Question

So I've been trying to calculate $$\sum_{a=2}^{\infty}\frac{2^{a}}{a\cdot a!}$$ but I'm stuck. The numerical value comes to be around 1.68 but I need a closed form expression. Any ideas?

Answer 1

As already remarked, there is no simple closed form since your series essentially is a value of the exponential integral: $$ \sum_{n\geq 2}\frac{2^n}{n\cdot n!}=\sum_{n\geq 2}\frac{2^n}{n!}\int_{0}^{1}x^{n-1}\,dx=\int_{0}^{1}\sum_{n\geq 2}\frac{(2x)^n}{n!}\cdot\frac{dx}{x}=-2+\int_{0}^{2}\frac{e^{x}-1}{x}\,dx.$$ On the other hand the RHS equals $$ -2+\int_{0}^{1}\frac{e^{2-2x}-1}{1-x}\,dx $$ and by approximating $e^{-2x}$ over $(0,1)$ with $\frac{1-x}{1+x}+\frac{x^2}{e^2}$ we already have that the original series is close to $e^2\log(2)-\frac{7}{2}$. Simpson's rule leads to the approximation $-\frac{19}{6}+\frac{4 e}{3}+\frac{e^2}{6}$ whose absolute error is already less than $6\cdot 10^{-3}$.

Answer 2

Since $e^x=\sum_{a=0}^\infty x^a/a!$, we have $$(e^x-1-x)/x=\sum_{a=2}^\infty \frac{x^{a-1}}{a!}.$$Now, integrate both sides from $0$ to $t$ with respect to $x$. Letting $f(x)$ be an antiderivative of $(e^x-1-x)/x$ (which I leave for you to find), you get $$ f(t)-f(0)=\sum_{a=2}^\infty \frac{t^a}{a\cdot a!} $$ Now plug in $t=2$ to both sides.