Series involving factorials How would one go about proving
$$\int_{0}^1\frac{e^x-1}{x/2}\ dx=\sum_{n=0}^\infty\frac{1}{\binom{n+2}{2}}\frac{1}{n!}(0!+1!+2!+3!+...+n!)$$
 A: Let 
$$f(x)=\sum_{n=0}^{\infty}  \frac{1} {(n+2)!}(0!+1!+2!+3!+...+n!)x^{n+2}$$
Then we obtain a differential equation
$$f'(x)=f(x)-\log(1-x)$$
This solves to
$$
f(x)=e^x\int_0^x (-e^{-t}\log (1-t))dt$$
The value that you want to find is $2f(1)$, so plug in $x=1$, 
$$
f(1)=e\int_0^1 (-e^{-t}\log (1-t))dt$$
A substitution $t$ to $1-t$, we have
$$
f(1)=\int_0^1 (-e^t\log t)dt$$
Then integration by part gives
$$
f(1)=-\log t (e^t -1)|_0^1+\int_0^1\frac{e^t-1}{t}dt$$
Thus
$$2f(1)=\int_0^1\frac{e^t-1}{t/2}dt.$$
This is the desired formula 
A: I had a direct approach. However I got stuck at the last step. If someone can check the steps and suggest a way to proceed, I would appreciate it:
Observe that $$\frac{e^x - 1}{x/2} = \frac{2}{x}\sum_{k=1}^\infty \frac{x^k}{k!}$$
Since x is in the nonnegative range, by Monotone convergence theorem, we have 
$$\int_{0}^{1}\frac{e^x - 1}{x/2}dx = 2\sum_{k=1}^\infty \int_{0}^{1}\frac{x^{k-1}}{k!}dx$$
$$= 2\sum_{k=1}^\infty \frac{1}{k!k}$$
$$= \sum_{k=0}^\infty \frac{2!}{(k+1)!(k+1)} =  \sum_{k=0}^\infty \frac{2!(k+2)}{(k+1)!(k+1)(k+2)}$$
$$= \sum_{k=0}^\infty \frac{2!}{(k+2)!}\frac{k+2}{k+1}$$
$$ = \sum_{k=0}^\infty \frac{1}{\binom{k+2}{2}k!}\frac{k+2}{k+1}$$
