f(x)=5/(5+x^2) ,c=0 Find a power series for the function, centered at c and determine the interval of convergence.

If x is a variable, then an infinite series of the form sum_(n=0)^ooa_n(x-c)^n=a_0+a_1(x-c)+a_2(x-c)^2+.......+a_n(x-c)^n+...... is a power series centered at x=c, where c is a constant.
Given f(x)=5/(5+x^2), c=0
Let's write f(x) in the form a/(1-r)
f(x)=5/(5(1+x^2/5))
f(x)=1/(1+x^2/5)
f(x)=1/(1-(-x^2/5))
which implies that a=1 and r=-x^2/5
So power series for f(x)=sum_(n=0)^ooar^n
=sum_(n=0)^oo(-x^2/5)^n
=sum_(n=0)^oo(-1)^n(x^2)^n/5^n
=sum_(n=0)^oo(-1)^nx^(2n)/5^n
This power series converges when |r|<1
=>|-x^2/5|<1
|x^2|<5
|x|Interval of convergence is (-sqrt5,sqrt5)