Dear All,
I am looking for the asymptotic behaviour of the sequence $a_n$ defined recursively by $a_1=1$ and $$ a_{n+1} = \frac{1}{n+1} \; \sum_{i=1}^n a_i \cdot \frac{1}{\Gamma\left(\frac{n-i}{2}+1\right)} . $$ The sequence $a_n$ converges to 0, but I need to know how fast it converges, i.e., I would like to have a simple equivalent. If $$ f(x) = \sum_{i>0} a_i x^i \; \; \mbox{ and } \;\; g(x) = \sum_{i\geq 0} \frac{1}{\Gamma\left(\frac{i}{2}+1\right)} x^i $$ then the above recursion translates into $$ f'(x) = f(x) g(x) +1. $$ The function $g$ can be expressed as a simple combination of the functions $\exp(x^2)$ and $erf(x)$. Based on the differential equation, one can compute a closed formula for $f$ and its coefficients, but the expression is quite ugly (due to the presence of integrals of $\exp(x^2)$ and $erf(x)$) and I was not able to derive the speed of convergence from this approach.
The closed expression for $f$ shows that $f$ is an entire function and therefore $a_n = O(\varepsilon^n)$ for all $\varepsilon > 0$.
An idea, anyone?
