Prove that sum is finite

Let $j \in \mathbb{N}$. Set $$ a_j^{(1)}=a_j:=\sum_{i=0}^j\frac{(-1)^{j-i}}{i!6^i(2(j-i)+1)!} $$ and $a_j^{(l+1)}=\sum_{i=0}^ja_ia_{j-i}^{(l)}$.

Please help me to prove that the following sum is finite $$ \sum_{j=1}^{\infty}j!\, a_j^{(l)} $$

Thank you.

Solution 1:

Define the generating functions for $a_n^{(l)}$ like so $$ g_l(z) = \sum_{n\geq0} a_n^{(l)}z^n$$ Then from the recurrence relations that you wrote down, it is easy to deduce that $$g_l(z) = \alpha(z)^l \beta(z)^l, \qquad \alpha(z) = e^{z/6}, \qquad \beta(z) = \frac{\sin\sqrt{z}}{\sqrt{z}}. $$

Write the sum that you have in the form $$ \sum_{n\geq0} a_n^{(l)}n! z^n = \sum_{n\geq0} \int_0^\infty dt\, e^{-t}(tz)^n a_n^{(l)} = \int_0^\infty e^{-t} g(tz). $$ If your sum converges absolutely, then this integral must converge also and have the same value, and if the integral converges (if it does it will converge absolutely), then the sum will converge. However, asymptotically $$ g(tz) = \Theta(e^{t z l /6} (tz)^{-l/2}), $$ so the integral will have an exponential term of the form $e^{(-1+zl/6)t}$. For the integral to converge the real part of $-1+zl/6$ must be negative, so the sum will converge if $$ \Re(z) \leq 6/l. $$

Setting $z=1$ to get your desired sum, it will converge absolutely when $l\leq 6$.