From the binomial expansion of #(1-x)^{-1/2}# we get
#(1-x)^{-1/2} = 1+(-1/2)(-x)#
#qquadqquadqquadqquad quad +1/(2!)(-1/2)(-3/2)(-x)^2+...#
#qquadqquadqquadqquad =1+1/2 x+1/(2!)* 1/2*3/2x^2+...#
we see that the coefficient of #x^n,\ \ n>=1# is
#1/(n!) *1/2*3/2*5/2*...*(2n-1)/2#
#=1/(n!) (1*3*5*...(2n-1))/2^n#
#= 1/(n!) (1*2*3*...*(2n-1)*(2n))/(2^n*2*4*6*...(2n))#
#= ((2n)!)/(2^{2n}(n!)^2)= 1/4^n((2n),(n))#
Thus
#(1-x)^{-1/2} = 1+sum_{n=1}^oo (x/4)^n((2n),(n)) = sum_{n=0}^oo (x/4)^n((2n),(n))#
and so
#sum_{n=1}^oo x^n((2n),(n)) = (1-4x)^{-1/2}#
So, our sum
#sum_{n=1}^oo 1/10^n((2n),(n)) = (1-4times 1/10)^{-1/2}=sqrt(5/3)#
