What is the interval of convergence of #sum_{k=0}^oo 2^(k) / ((2k)!* x^(k)) #?

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Answer 1 · 2016-07-20T20:18:07

#sum_{k=0}^oo 2^(k) / ((2k)!* x^(k))# converges #forall x in RR, x ne 0#

We know that

#e^x=sum_{k=0}^oox^k/(k!)# and also
#(e^x+e^-x)/2 = cosh(x) = sum_{k=0}^oox^{2k}/(2k!)#

but

#sum_{k=0}^oo 2^(k) / ((2k)!* z^(k))=sum_{k=0}^oo(2/z)^k/(2k!)=sum_{k=0}^oo(sqrt(2/z))^{2k}/(2k!)#

so for #x > 0#

#sum_{k=0}^oo(sqrt(2/x))^{2k}/(2k!)=cosh(sqrt(2/x))#

Analogous considerations can be stated for #x < 0#