Original problem is to determine asymptotic behavior of ${a_i}\left( t \right)$ as $t \to \infty $ given by recurrence relations
${a_1}\left( 0 \right) = 1$
${a_1}\left( t \right) = \frac{{2t + 1}}{{2t}}{a_1}\left( {t - 1} \right)$
${a_i}\left( t \right) = \frac{{2t + 1}}{{2t}}{a_i}\left( {t - 1} \right) + \frac{1}{{2t}}{a_{i - 1}}\left( {t - 1} \right)$
for $i \in \mathbb{N}$.
My attempt at solution:
Generating function for ${a_i}$ is $\displaystyle{F_i}\left( x \right) = \frac{{{{\left( {1 - x} \right)}^{ - \frac{3} {2}}}{{\ln }^{i - 1}}\frac{1} {{1 - x}}}} {{\left( {2i - 2} \right)!!}}$.
If we rewrite recurrence relations in a different way:
${a_1}\left( t \right) - {a_1}\left( {t - 1} \right) = \frac{1} {{2t}}{a_1}\left( {t - 1} \right)$
${a_i}\left( t \right) - {a_i}\left( {t - 1} \right) = \frac{1} {{2t}}{a_i}\left( {t - 1} \right) + \frac{1} {{2t}}{a_{i - 1}}\left( {t - 1} \right)$
And then pretend that difference operator is differential operator
${a_1}^\prime \left( t \right) = \frac{1} {{2t}}{a_1}\left( t \right)$
${a_i}^\prime \left( t \right) = \frac{1} {{2t}}{a_i}\left( t \right) + \frac{1} {{2t}}{a_{i - 1}}\left( t \right)$
by solving a system of differential equations, we get ${a_i}\left( t \right) = \frac{{\sqrt {1 + t} {{\ln }^{i - 1}}\left( {1 + t} \right)}} {{\left( {2i - 2} \right)!!}}$. Now we define ${G_i} = \sum\limits_{t = 0}^\infty {\frac{{\sqrt {1 + t} {{\ln }^{i - 1}}\left( {1 + t} \right)}} {{\left( {2i - 2} \right)!!}}{x^t}} $.
It turns out that (see http://www.math.upenn.edu/~pemantle/papers/twenty.pdf , page 210-211) if ${L_i} = \frac{{{G_i}\left( x \right)}} {{{F_i}\left( x \right)}} = \mathop {\lim }\limits_{x \to 1 - } \frac{{\sum\limits_{t = 1}^\infty {\sqrt t {x^t}{{\ln }^{i - 1}}t} }} {{x{{\left( {1 - x} \right)}^{ - \frac{3} {2}}}{{\ln }^{i - 1}}\frac{1} {{1 - x}}}} = \mathop {\lim }\limits_{x \to 1 - } \frac{{{{\left( {1 - x} \right)}^{\frac{3} {2}}}\sum\limits_{k = 1}^\infty {\sqrt k {x^k}{{\ln }^{i - 1}}k} }} {{{{\ln }^{i - 1}}\frac{1} {{1 - x}}}}$ exists and is nonzero, then $\displaystyle\mathop {\lim }\limits_{t \to + \infty } {a_i}\left( t \right) \cdot \frac{{\left( {2i - 2} \right)!! \cdot {L_i}}} {{\sqrt {t + 1} {{\ln }^{i - 1}}\left( {t + 1} \right)}} = 1$. For several values of ${L_i}$ this seems to be true, which strikes me as unexpected.
For $i = 1$ wolfram's mathematica returns ${L_1} = \frac{{\sqrt \pi }}{2}$, but can not evaluate other expressions. However, they are around 0.9 and rising as $i$ rises.
My question is how to analytically evaluate ${L_i}$, or alternatively, disregarding the previous arguments, how to determine asymptotic behavior of a sequence ${a_i}\left( t \right)$?
More generally, is it usual that the solutions of difference and corresponding differential equation (given by replacing difference operator with differential operator) have the same asymptotic behavior up to a multiplicative constant? If that is the case and such behavior is well understood, are there any textbooks about it?