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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

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