How do you find the power series for $f(x)=arctanx/x$ and determine its radius of convergence?

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Answer 1 · 2017-12-09T15:49:49

$arctanx/x = sum_(n=0)^oo (-1)^n x^(2n)/(2n+1)$

converging for $absx < 1$ .

Start the sum of the geometric series:

$sum_(n=0)^oo q^n = 1/(1-q)$ for $absq < 1$

posing: $q = -x^2$ , we have:

$sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^n x^(2n) = 1/(1+x^2)$ for $absx < 1$

In the interval of convergence we can integrate term by term:

$int dx/(1+x^2) = sum_(n=0)^oo int (-1)^n x^(2n) = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) + C$

As:

$int dx/(1+x^2) = arctanx +C$

we have:

$arctanx = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) + C$

as $arctan0 = 0$ we can see that $C=0$ , and dividing by $x$ term by term:

$arctanx/x = sum_(n=0)^oo (-1)^n x^(2n)/(2n+1)$

converging for $absx < 1$ .

How do you find the power series for f(x)=arctanx/xf(x)=arctanx/x and determine its radius of convergence?