Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 16 down vote favorite
6

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.

share|improve this question
1  
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... – Michael Dec 10 '12 at 2:10
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 ? – mike Dec 13 '12 at 17:14
@mike: Your idea sounds interesting. Could you please elaborate. Thank you. – Michael Dec 13 '12 at 23:46
@ mike: do you have suggestions where I can read about generating functions. Thank you. – Alex Feb 5 at 15:57
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 $1$? – Enjoys Math yesterday
show 1 more comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.