The question asks to use the multi-section technique to determine if $$\sum_{n>=0} (a^n)/(4n +1)!$$

converges, and to provide a finite expression for the exact value of the series.

The multi section technique: let $\omega = e^{2πi/r}$ and $A(x) = \sum_na_nx^n$ then $$[A(x) + A(\omega x) + A(\omega^2 x) ... A(\omega^{r-1}x)]1/r = \sum_j a_{rj}x^{rj}$$ So from here, I've been trying to equate $$\sum_j a_{rj}x^{rj} = \sum_{n>=0} (a^n)/(4n +1)!$$ looking to split the right hand side up into a sequence of A($\omega^k$x)'s so that I can use the multisection technique to break up into smaller more manageable parts that may simplify into a finite expression. Unfortunately, I'm not sure how to split it up in a way that gives me any desirable result. Any hints would be a big help!

  • 1
    $\begingroup$ It seems that $r=4$ is the right choice, i.e. $[1,i,-1,-i]$ where $i=\sqrt{-1}$. I say this from looking at maple's answer to the sum, which has 3 terms, but a sine term is weighted twice, likely from combining exponentials of type $e^{it}$. $\endgroup$ – coffeemath Feb 15 '13 at 18:01

The series $\sum_{n\ge 0} a^n/(4n +1)!$ is itself a section of an exponential: \begin{eqnarray*} e^x + i^{-1} e^{ix} - e^{-x} - i^{-1} e^{-ix}&=& \sum_{m\ge 0} \frac{x^m}{m!} (1 + i^{m-1} + (-1)^{m-1} + (-i)^{m-1})\\ &=&4 \sum_{n\ge 0} \frac{x^{4n+1}}{(4n+1)!} \end{eqnarray*} so \begin{eqnarray*} \sum_{n\ge 0} \frac{a^n}{(4n+1)!}&=&\frac{1}{4\sqrt[4]{a}} (e^{\sqrt[4]{a}}-ie^{i\sqrt[4]{a}} -e^{-\sqrt[4]{a}}+ie^{-i\sqrt[4]{a}})\\ &=& \frac{1}{2\sqrt[4]{a}} (\sinh \sqrt[4]{a} + \sin \sqrt[4]{a}). \end{eqnarray*}


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