This may be a simple question, but I was slightly confused. I was looking at the second line $S_n(x)=1-x^{n+1}/(1-x)$. I was confused how they derived this. I know the infinite sum of a geometric series is $1/(1-x)$. I just can't figure out how the partial sums, $S_n(x)$, have $1-x^{n+1}$ on the numerator. How was this derived?
Thank you.
Example 5.20.
The geometric series
$$
\sum_{n=0}^\infty x^n
= 1 + x + x^2 + x^3 + \dotsb
$$
has partial sums
$$
S_n(x)
= \sum_{k=0}^n x^k
= \frac{1 – x^{n+1}}{1 – x} \cdotp
$$
Thus, $S_n(x) \to 1/(1-x)$ as $n \to \infty$ if $|x| < 1$ and diverges if $|x| \geq 1$, meaning that
$$
\sum_{n=0}^\infty x^n
= \frac{1}{1-x}
\qquad
\text{pointwise on $(-1,1)$}.
$$
(Original image here.)
Best Answer
It's from the sum of a (finite) geometric series. But you can derive it from first principles.
$$S_n(x) = 1 + x + x^2 + \dotsb + x^n$$
$$xS_n(x) = x + x^2 + x^3 + \dotsb + x^{n+1}$$
Subtracting the second from the first (and noting the telescoping nature, which I'm making explicit here),
$$(1-x)S_n(x) = 1 - x + x - x^2 + x^2 + \dotsb - x^n + x^n - x^{n+1} = 1- x^{n+1}.$$
Rearranging,
$$S_n(x) = \frac{1-x^{n+1}}{1-x}.$$