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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... –  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 '13 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 May 12 '13 at 4:43
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1 Answer

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 $l\leq 6$.

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