Start the sum of the geometric series:
sum_(n=0)^oo q^n = 1/(1-q)∞∑n=0qn=11−q for absq < 1|q|<1
posing: q = -x^2q=−x2, we have:
sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^n x^(2n) = 1/(1+x^2)∞∑n=0(−x2)n=∞∑n=0(−1)nx2n=11+x2 for absx < 1|x|<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∫dx1+x2=∞∑n=0∫(−1)nx2n=∞∑n=0(−1)nx2n+12n+1+C
As:
int dx/(1+x^2) = arctanx +C∫dx1+x2=arctanx+C
we have:
arctanx = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) + Carctanx=∞∑n=0(−1)nx2n+12n+1+C
as arctan0 = 0arctan0=0 we can see that C=0C=0, and dividing by xx term by term:
arctanx/x = sum_(n=0)^oo (-1)^n x^(2n)/(2n+1)arctanxx=∞∑n=0(−1)nx2n2n+1
converging for absx < 1|x|<1.
