What is the interval of convergence of the Taylor series of #f(x)=1/(1-x)#?

1 Answer
Mar 28, 2018

The Taylor Series of #f(x)=1/(1-x)# centered at #a=0# is

#sum_(n=0)^oox^n=1+x+x^2+x^3+cdots#

which converges for #-1##<##x<1#

Explanation:

The general formula for the Taylor Series of #f(x)# centred at #a# is

#sum_(n=0)^oo (f^((n))(a))/(n!)(x-a)^n#

Let's find the general form for the #n#th derivative of #f(x)=1/(1-x)#

#f'(x)=1/(1-x)^2#

#f''(x)=2/(1-x)^3#

#f'''(x)=(2*3)/(1-x)^4#

We can see the pattern emerging.

The formula appears to be

#f^((n))(x)=(n!)/(1-x)^(n+1)#

For simplicity, let's centre our Taylor Series at #a=0#

Then #f^((n))(a)# is:

#f^((n))(0)=(n!)/(1-0)^(n+1)=n!#

and the Taylor Series becomes:

#sum_(n=0)^oo (n!)/(n!)(x-0)^n=sum_(n=0)^oox^n#

Now that we have our series, let's use the ratio test to check for convergence.

#lim_(n->oo)abs(x^(n+1)/x^n)=abs(x)#

Therefore the series converges for #abs(x)<1#

Which can be expressed as #-1##<##x<1#

Let's check the endpoints of this interval.

Check #x=-1#

#rArrsum_(n=0)^oo(-1)^n=1-1+1-1+cdots-cdots#

#rarr# Does not converge.

Check #x=1#

#rArrsum_(n=0)^oo1^n=1+1+1+1+cdots#

#rarr# Diverges to #+oo#

So the final answer is:

#sum_(n=0)^oox^n# converges for #-1##<##x<1#