I have this sequence
am=∞∑k=0(k+mm)14k(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
a0=cosh(12)a1=sinh(12)a2=12(cosh(12)−2sinh(12))⋮
but cosh(1/2) and sinh(1/2) keeps showing up for a3, a4, ⋯
Could anyone find a general expression for am involving cosh(1/2) and sinh(1/2).
I'd apprecaite any help, thanks.
Answer
am=∑k≥0(k+m)!4−km!k!(2k+2m)!=4mm!⋅dmdxm∑k≥0(x/4)k+m(2k+2m)!|x=1
and now you may recognize in the last series part of the Taylor series of cosh(√x/2).
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