A power series is also called Taylor series. Taylor series are represented by #sum_(n=0)^oo(f^n(a)(x-a)^n)/(n!)#, where the series is approximated around a value #a#.

Now, #f(x)=ln(x^2+1)# looks ridiculously similar to the function #g(x)=ln(1+x)#, and for that function there is a well known Taylor expansion centered around #a#, which is:

#g(x)=ln(1+a)+sum_(n=1)^oo((x-a)^n(-1)^(n+1))/(n(1+a)^n)#

Letting #h(x)=x^2#, and recognising that #f(x)=ln(x^2+1)-=g(h(x))#, we can substitute #h(x)# into the series expansion for #g(x)#.

#f(x)=g(h(x))=ln(1+h(a))+sum_(n=1)^oo((h(x)-a)^n(-1)^(n+1))/(n(1+a)^n)=ln(1+a^2)+sum_(n=1)^oo((x^2-a)^n(-1)^(n+1))/(n(1+a)^n)#

**Therefore, the Taylor series (power series) approximation for #f(x)=ln(x^2+1)# centered around #x=a# is #ln(1+a^2)+sum_(n=1)^oo((x^2-a)^n(-1)^(n+1))/(n(1+a)^n)#**

Now, we find its radius of convergence by applying the ratio test. The ratio test works by finding the ratio between the #(n+1)th# term and the #nth# term as #n# tends to #oo#. As a condition for convergence is that the ratio #r# has to satisfy #absr<1# , we can limit the #x# for which the series will converge. I will show you how below:

Applying the ratio test:

#lim_(k->oo)abs(u_(k+1)/u_k)
=lim_(k->oo)abs((((x^2-a)^(k+1)(-1)^(k+1+1))/((k+1)(1+a)^(k+1)))/(((x^2-a)^k(-1)^(k+1))/(k(1+a)^k)))
=lim_(k->oo)abs(((x^2-a)^(k+1))/((k+1)(1+a)^(k+1))*(k(1+a)^k)/((x^2-a)^k))
=lim_(k->oo)abs((x^2-a)/(1+a)*k/(k+1))
=abs((x^2-a)/(1+a))*lim_(k->oo)abs(k/(k+1))
=abs((x^2-a)/(1+a))*1#

For the series to converge, the absolute ratio has to be less than 1, i.e.

#abs((x^2-a)/(1+a))<1#

#abs(x^2-a)<1+a#

#abs(x^2)<1+2a#

#absx< sqrt(1+2a)#

**The radius of convergence is therefore #sqrt(1+2a)#, which as we can see is dependent on the center of the series #a#.**