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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.

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    Its an interesting question and I am confused how to solve it. I was trying to attempt using Cauchy-Schwarz inequality, but the result was weaker than one needed. Also, some estimates using combinatorics did not give any reasonable result...2012-12-10
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    I think $a_j$ itself has the form of a convolution, and one can find generating function $\sum a_j x^j$, and therefore that of all the $a_j^l$, but I (could be wrong and) wonder if the poser knows all this ?2012-12-13
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    @mike: Your idea sounds interesting. Could you please elaborate. Thank you.2012-12-13
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    @ mike: do you have suggestions where I can read about generating functions. Thank you.2013-02-05
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    What do you mean $a_j^{(l+1)}=\sum_{i=0}^ja_ia_{j-i}^{(l)}$ do you mean there is a coefficient $a_i$ for each term that depends on $j$? or do did you mean $a_i^{(l)}$?? Also, for the latter, starting at $0$ in the summation gives a self-referencing formula so did you mean from $1$?2013-05-12
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    My last sentence above is wrong2013-05-12

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