Finding a closed form expression of a sequence that is defined recursively via a definite integral Consider the following series function that is defined recursively by the following definite integral
$$
f_n(x) = \int_0^x u^n f_{n-1}(u) \, \mathrm{d}u  \qquad\qquad (n \ge 1) \, ,
$$
with $f_0 (x) = \operatorname{erf}(x)$ being the error function.
Then, let us define for $n \ge 0$ the following sequence:
$$
u_n = \lim_{x\to 0} x^{-\frac{(n+1)(n+2)}{2}} f_n(x) \, .
$$
It can readily be checked that $0 < u_n < \infty$.
My questions are:

*

*Is there a way to find out the general term of the above number sequence?


*Is the series $\sum_{n=0}^{\infty} u_n$ convergent? If so, what is the limit of this series?
Any hint of help is highly appreciated and desirable.
Thanks and best,
Daddy

@Szeto has been faster than myself.
The $f_n(x)$ are not very difficult to compute. Expanding them as Taylor series around $x=0$, the very first term corresponds to the sequence
$$\left\{\frac{2 x}{\sqrt{\pi }},\frac{2 x^3}{3 \sqrt{\pi }},\frac{x^6}{9 \sqrt{\pi
   }},\frac{x^{10}}{90 \sqrt{\pi }},\frac{x^{15}}{1350 \sqrt{\pi
   }},\frac{x^{21}}{28350 \sqrt{\pi }},\frac{x^{28}}{793800 \sqrt{\pi }},\cdots\right\}$$
Defining, as @Szeto proposed, $v_n=\frac {u_n}{u_0}$ this generates the sequence
$$\left\{1,\frac{1}{3},\frac{1}{18},\frac{1}{180},\frac{1}{2700},
\frac{1}{56700},\cdots\right\}$$ The denominators correspond to sequence $A006472$ in $OEIS$
$$v_n=\frac{2^{n+1}}{(n+1)! (n+2)!}\implies u_n=\frac{1 }{\sqrt{\pi }}\frac{2^{n+2}}{(n+1)! (n+2)!}$$
$$g(t)=\frac{1 }{\sqrt{\pi }}\sum_{n=0}^\infty \frac{2^{n+2}}{(n+1)! (n+2)!} t^n=\frac{\sqrt{2} I_1\left(2 \sqrt{2} \sqrt{t}\right)-2 \sqrt{t}}{\sqrt{\pi }\, t^{3/2}}$$
$$\sum_{n=0}^\infty u_n=g(1)=\frac{\sqrt{2} I_1\left(2 \sqrt{2}\right)-2}{\sqrt{\pi }} \approx 1.5739$$
