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
 A: Apparently you made some index mistakes...
If $$f_n(x)=u_n x^{k_n}+O\left(x^{k_n+1}\right)$$ then it can be easily seen
$$k_{n+1}=k_n+n+2$$
$$u_{n+1}=\frac{u_n}{k_n+n+2}$$
Assuming that you are using the unnormalised error function $\text{erf}(x)=x+O(x^3)$, we have the initial conditions $k_0=1$, $u_0=1$.
Solving the first recurrence relation we get $$k_n=\frac{(n+1)(n+2)}2$$
Obviously, the second recurrence relation gives $$u_n=\prod^{n-1}_{i=0}\frac1{k_i+i+2}=\prod^n_{i=1}\frac2{(i+1)(i+2)}=\frac{2^{n+1}}{(n+1)!(n+2)!}$$
Doing a little algebra one arrives at $$\sum^\infty_{n=0}u_n=\frac{I_1\left(2\sqrt2\right)}{\sqrt2}-1=1.394833...$$ where $I_1$ is the first order modified Bessel function of the first kind.
A: @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 $OEIS$
$$v_n=\frac{2^{n+1}}{(n+1)! (n+2)!}\implies u_n=\frac{1 }{\sqrt{\pi }}\frac{2^{n+2}}{(n+1)! (n+2)!}$$
$$g(t)=\frac{1 }{\sqrt{\pi }}\sum_{n=0}^\infty \frac{2^{n+2}}{(n+1)! (n+2)!} t^n=\frac{\sqrt{2} I_1\left(2 \sqrt{2} \sqrt{t}\right)-2 \sqrt{t}}{\sqrt{\pi }\, t^{3/2}}$$
$$\sum_{n=0}^\infty u_n=g(1)=\frac{\sqrt{2} I_1\left(2 \sqrt{2}\right)-2}{\sqrt{\pi }} \approx 1.5739$$
