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 $x < 0$

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