How do you find a power series converging to #f(x)=sinx/x# and determine the radius of convergence?

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Answer 1 · 2017-02-15T14:36:11

#sinx/x = sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!)#

with radius of convergence #R=oo#.

Start from the MacLaurin series for #sin x#:

#sinx = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)#

We can divide by #x# term by term:

#sinx/x = sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!)#

and determine the radius of convergence using the ratio test:

#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs( (x^(2(n+1))/((2(n+1)+1)!))/( x^(2n)/((2n+1)!)#

#lim_(n->oo) abs(a_(n+1)/a_n) = abs((x^(2n+2))/x^(2n)) ((2n+1)!)/((2n+3)!)#

#lim_(n->oo) abs(a_(n+1)/a_n) = x^2/((2n+3)(2n+2)) = 0#

So the series is convergent for every #x#.