Question

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

Answer 1

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

with radius of convergence `R=oo`.

Explanation:

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`.