## Proposition 8.4.10: The Geometric Series

For the geometric series we can actually compute the remainder in Taylor's
theorem exactly. Recall that we have previously shown that if we define
*S _{n}(x) = 1 + x + x^{2} + ... + x^{n}*
we can find a closed form for

*S*by multiplying

_{n}(x)*S*by

_{n}(x)*x*and then subtracting

*S*to get:

_{n}(x) - x S_{n}(x)S_{n}(x) =^{1-xn+1}/_{1-x}

or equivalently

1 + x + x^{2}+ ... + x^{n}=^{1 - xn+1}/_{1 - x}=^{1}/_{1-x}-^{xn+1}/_{1-x}

We can rearrange the terms to get:

^{1}/_{1-x}- (1 + x + x^{2}+ ... + x^{n}) =^{xn+1}/_{1-x}

But this means that we have (with *c = 0*)

|f(x) - | = |R_{n+1}(x)| = |^{xn+1}/_{1 - x}|

Thus, we have found the remainder exactly! And since *|x| < 1*
it is now clear that this remainder converges to zero, so that by Taylor's
theorem the Taylor series converges to the function, as required.