Tuesday, 26 April 2016

sequences and series - Looking for a closed form for am=suminftyk=0binomk+mmfrac14k(2(k+m))!



I have this sequence
a_m =\sum_{k=0}^{\infty}\binom{k+m}{m}\frac{1}{4^{k}(2(k+m))!}
and there seems to exist a patern arising when it is evaluated by WA. It involves \cosh(1/2) and \sinh(1/2). Bellow are the first
\begin{align*} a_0 &= \cosh\left(\frac{1}{2}\right)\\ a_1 &= \sinh\left(\frac{1}{2}\right)\\ a_2 &= \frac{1}{2}\left(\cosh\left(\frac{1}{2}\right )-2\sinh\left(\frac{1}{2} \right ) \right )\\ &\vdots\\ \end{align*}
but \cosh(1/2) and \sinh(1/2) keeps showing up for a_3, a_4, \cdots



Could anyone find a general expression for a_m involving \cosh(1/2) and \sinh(1/2).




I'd apprecaite any help, thanks.


Answer



a_m = \sum_{k\geq 0}\frac{(k+m)! 4^{-k}}{m! k! (2k+2m)!}=\frac{4^m}{m!}\cdot\left.\frac{d^m}{dx^m}\sum_{k\geq 0}\frac{(x/4)^{k+m}}{(2k+2m)!}\right|_{x=1}
and now you may recognize in the last series part of the Taylor series of \cosh(\sqrt{x}/2).


No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...