# Prove that sum is finite with the help of generating function

Please help me to prove that the following sum is finite $$\sum_{j=2l-2}^{\infty}j!\, a_j^{(l)},$$ here the generating function of $\displaystyle{a_j^{(l)}}$ is $\displaystyle{\left(\frac{e^{x/6}\sin{\sqrt x}}{\sqrt x}\right)^l}.$

I.e., we should show convergence of $$\sum_{j=2l-2}^{\infty}j!\, \left[\left(\frac{e^{x/6}\sin{\sqrt x}}{\sqrt x}\right)^l\right]^{(j)},$$ where one should take the $(j)$-th derivative at $x=0$, i.e. limit when $x$ approach to zero.  This question was posted by another user under the different contekst here Prove that sum is finite and someone gave an idea that maybe generating function would be of help.

I was trying to study a generating functions (its not my field) but I did not succeed. So I've asked this: Find generating function $\sum_{j}a_jx^j$ so that allows to find all of $a_j^{\ell}$. (see Construct a generating function for the components of a sum).

Now, I am trying to use this generating function, but I am still far away from the solution. As I understand correctly, now instead of $\sum_{j=1}^{\infty}j!\, a_j^{(l)}$ I have series with the $\ell$-th derivative of a generating function. But is it really simplify the solution?

Thank you.

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Please reread Antonio's answer in your $2^{nd}$ link. If you understand his answer, then you will immediately see the generation function for $a_j^{(l)}$ is given by $\left(\frac{e^{x/6}\sin{\sqrt{x}}}{\sqrt{x}}\right)^l$. – achille hui Feb 7 '13 at 17:31
No, I have hot whats the generating function. My question is how to show that series is convergentn... – Michael Feb 7 '13 at 20:02