Question

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

Answer 1

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`