We can develop the function in MacLaurin series:
#f(x) = sum_(n=0)^oo (f^((n))(0))/(n!)x^n#
Evaluating the successive derivatives:
#f'(x) = -1/2(1-x)^(-3/2)#
#f''(x) = (-1/2)(-3/2)(1-x)^(-5/2)#
#...#
we can construct by induction the general form of the #n#-th derivative:
#f^((n))(x) = (-1)^n ((2n+1)!!)/2^n(1-x)^((2n-1)/2)#
so that:
#f^((n))(0) = (-1)^n ((2n+1)!!)/2^n#
and then:
#1/sqrt(1-x) = sum_(n=0)^oo (-1)^n ((2n+1)!!)/ (2^n n!) x^n#
The coefficients of the series are called generalized binomial coefficients:
#((k),(n)) = underbrace( k (k-1) (k-2) * ... * (k-n+1))_(n " times") = (Gamma(k))/(Gamma(n+1)Gamma(k-n+1))#
To determine the interval of convergence we can apply the ratio test:
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs ( (((2n+3)!!)/ (2^(n+1) (n+1)!) x^(n+1))/(((2n+1)!!)/ (2^n n!) x^n))#
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) abs( ((2n+3)!!)/ ((2n+1)!!) 2^n/2^(n+1) (n!)/((n+1)!) x^(n+1)/x^n)#
#lim_(n->oo) abs(a_(n+1)/a_n) = lim_(n->oo) (2n+3)/(2(n+1))abs( x) = absx#
So, the series is absolutely convergent for #x in (-1,1)# and not convergent in #(-oo,-1) uu (1,+oo)#
In the case where #absx = 1# we have that for #x=1# the function is not defined, and for #x = -1# the series becomes:
#sum_(n=0)^oo ((2n+1)!!)/ (2^n n!) #
if we look at the general term of this series we have:
#a_n = ((2n+1) (2n-1) ... 3)/(2n(2n-2)...2) = ((2n+1)/(2n))((2n-1)/(2n-2))...(3/2)#
so: #lim_(n->oo) != 0# and the series is not convergent.
