We start by working out a taylor series for #ln(1+x)#. I will be expanding around #0#, so it will be a Maclaurin series.
The general formula for a Maclaurin series is:
#f(x)=sum_(n=0)^oof^n(0)/(n!)x^n#
This means we need to work out the nth derivative of #ln(1+x)#. Let's start by taking some derivatives and see what values they produce at #x=0#:
#f(x)=ln(1+x)-> 0#
#f'(x)=1/(1+x)-> 1#
#f''(x)=-1/(1+x)^2-> -1#
#f^3(x)=2/(1+x)^3-> 2#
#f^4(x)=-6/(x+1)^4-> -3#
#f^5(x)=24/(x+1)^5-> 24#
#color(white)(.....................)⋮#
#f^n(x)=(-1)^(n+1)((n-1)!)/(x+1)^(n)-> (-1)^(n+1)(n-1)!#
We can now plug this into the Maclaurin expansion (note that we ignore the first term, since it is #0#):
#ln(1+x)=sum_(n=1)^oo((-1)^(n+1)(n-1)!)/(n!)x^n#
This simplifies to:
#ln(1+x)=sum_(n=1)^oo(-1)^(n+1)/nx^n#
Now that we have a series for #ln(1+x)#, we can replace all the #x#'s with #x^2# to get a series for #ln(1+x^2)#:
#ln(1+x^2)=sum_(n=1)^oo(-1)^(n+1)/n(x^2)^n=#
#=sum_(n=1)^oo(-1)^(n+1)/nx^(2n)=x^2-x^4/2+x^6/3-x^8/4...#
which is the series we were looking for.
To test the convergence, we can use the ratio test:
#lim_(n->oo)-(x^(2(n+1))/(n+1))/(x^(2n)/n)=lim_(n->oo)-(x^2cancel(x^(2n)))/(n+1)*n/cancel(x^(2n))=#
#=lim_(n->oo)-(nx^2)/(n+1)=-x^2#
So, we need to look at when #|-x^2|<1#:
#|-x^2|<1#
#x^2<1#
#sqrt(x^2)< sqrt1#
#|x|<1#
which means the series is convergent when #|x|<1#
