Regards.. I would like to discuss about the convergence function of the series $$ \sum_{0}^{\infty} \: \frac{x^{n}}{n!(n+1)!}$$ If i write this series as function $f(x)$ and then integrate it, i would get : $$ F(x) =\sum_{0}^{\infty} \: \frac{x^{n+1}}{((n+1)!)^{2}} $$ But, i have not found any explicit function for this series.
I may try that the function $f(x)$ must be of the form $ f(x) = g(x) e^{x} $. Then writing $g(x)$ and $e^{x}$ as power series :
\begin{align*} f(x) &= \sum_{0}^{\infty} \: \frac{x^{n}}{n!(n+1)!} \\ f(x) &= \left( a_{0} + a_{1}x + a_{2} x^{2} + ... \right) \times (1 + x + \frac{x^{2} }{2!} + \frac{x^{3}}{3!} + ... ) \end{align*}
Then, for few terms, $x$ and $x^{2}$ coefficients are : $ (a_{0} + a_{1}) = \frac{1}{1!2!}$,
$ (a_{0} + a_{1} + \frac{a_{2}}{2}) = \frac{1}{2!3!}$. The result is $a_{2} = -5/6 $.
If I may have some analysis of this, but may not be the full answer, direction is better. Thanks.
This is related to the modified Bessel function $$I_\alpha(x) = \sum_{n=0}^\infty \frac{(x/2)^{2n+\alpha}}{n!\, \Gamma(n+\alpha+1)}.$$ More precisely, we can write the power series you're looking for as $$\sum_{n=0}^\infty \frac{x^n}{n!\,(n+1)!} = \frac1{\sqrt x} I_1(2\sqrt{x}).$$ So in particular there's no closed form in terms of elementary functions, but there are some nice differential equations and such.