# Partial Sums of Geometric Series

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.)

It's from the sum of a (finite) geometric series. But you can derive it from first principles.

$$S_n(x) = 1 + x + x^2 + ... + x^n$$

$$xS_n(x) = x + x^2 + x^3 + ... + x^{n+1}$$

Subtracting the second from the first (and noting the telescoping nature, which I'm making explicit here),

$$(1-x)S_n(x) = 1 - x + x - x^2 + x^2 + ... - x^n + x^n - x^{n+1} = 1- x^{n+1}$$

Rearranging,

$$S_n(x) = \frac{1-x^{n+1}}{1-x}$$

Observe that $$\frac{1}{x-1}(x^{k+1}-x^{k})=x^k\quad (x\neq 1)$$ whence $$\sum_{k=0}^n x^k=\sum_{k=0}^n\frac{1}{x-1}(x^{k+1}-x^{k})=\frac{1}{x-1}(x^{n+1}-1) =\frac{1-x^{n+1}}{1-x};\quad (x\neq 1)$$ since the sum telescopes.

• Got it, thanks. – rain Feb 11 '18 at 0:56

$S_n(x)=1+x+x^2+x^3+ . . .x^n=1+x+x^2+x^3+ . . .x^n +x^{n+1}-x^{n+1}=1-x^{n+1} + x(1 +x+x^2+x^3 . . .+x^n)=1-x^{n+1} +x S_(n)$

⇒ $(1-x)S_n(x)=1-x^{n+1}$

⇒ $S_n(x)=\frac{1-x^{n+1}}{1-x}$