# Partial Sums of Geometric Series

Mathematics Asked on January 5, 2022

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.)

For completeness let me add one, also very usual, argumentation. Partial sum can be derived from formula: $$a^{n}-b^{n}=(a-b)(a^{n-1} + ba^{n-2}+ cdots + b^{n-1})$$ Taking $$b=1$$ we obtain $$a^{n}-1=(a-1)(a^{n-1} + a^{n-2}+ cdots + 1) Rightarrow a^{n-1} + a^{n-2}+ cdots + 1 = frac{a^{n}-1}{a-1}$$

Answered by zkutch on January 5, 2022

Deepak's excellent answer is the standard argument. I will give (essentially) the same argument here, but a slightly different presentation. What I like about this argument, in comparison to Deepak's, is that it eliminate the ellipses and makes the computations a little more precise. There is a cost—I think that some of the intuition is lost, since we don't see the term-by-term cancelation—but I think that this is a price which can be paid without too much difficulty.

Given any real (or complex) number $$x$$ and any natural number $$n$$, let $$S_n(x)$$ denote the $$n$$-th partial sum of the series $$sum_{j=0}^{infty} x^j$$. That is, $$S_n(x) = sum_{j=0}^{n} x^j.$$ Observe that begin{align} xS_{n}(x) - S_{n}(x) &= xsum_{j=0}^{n} x^j - sum_{j=0}^{n} x^j \ &= sum_{j=0}^{n} x^{j+1} - sum_{j=0}^{n} x^j &&text{(distribution over finite sums)} \ &= sum_{k=1}^{n+1} x^{k} - sum_{j=0}^{n} x^j &&text{(CoV: let k=j+1)} \ &= left[ sum_{k=1}^{n} x^k + x^{n+1}right] - left[1 + sum_{j=1}^{n} x^j right] && text{(pull out a couple of terms)} \ &= x^{n+1} + color{red}{sum_{k=1}^{n} x^k} - color{red}{sum_{j=1}^{n} x^j} - 1 && text{(the red terms cancel)} \ &= x^{n+1} - 1. end{align} Supressing the intermediate steps, this reduces to begin{align} x S_n(x) - S_n(x) = x^{n+1} - 1 &implies (x-1)S_n(x) = x^{n+1} - 1 \ &implies S_n(x) = frac{x^{n+1}-1}{x-1} = frac{1-x^{n+1}}{1-x}, end{align} which is the claimed identity.

Another alternative to Deepak's appeal to telescoping sums is the following. Again, we start with begin{align} (x-1)S_n(x) &= xS_n(x) - S_n(x) \ &= left[ x + x^2 + x^3 + dotsb + x^n + x^{n+1} right] - left[ 1 + x + x^2 + x^3 + dotsb + x^n right]. end{align}

If we write this subtraction in the style that is taught in American elementary schools, it looks something like $$begin{array}{r} &&& color{red}{x} &+& color{blue}{x^2} &+& color{green}{x^3} &+& dotsb &+& color{orange}{x^{n}} &+& x^{n+1} \ -{quad} & 1 &+& color{red}{x} &+& color{blue}{x^2} &+& color{green}{x^3} &+& dotsb &+& color{orange}{x^{n}} \hline & -1 &+& color{red}{0} &+& color{blue}{0} &+& color{green}{0} &+& dotsb &+& color{orange}{0} &+& x^{n+1}. end{array}$$ Lining up "like terms" (that is, aligning terms with the same exponent) makes it a little easier to see where the cancelations are happening. The rest of the argument is identical.

Answered by Xander Henderson on January 5, 2022

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

⇒ $(1-x)S_n(x)=1-x^{n+1}$

⇒ $S_n(x)=frac{1-x^{n+1}}{1-x}$

Answered by sirous on January 5, 2022

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,

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

Answered by Deepak on January 5, 2022

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

Answered by Sri-Amirthan Theivendran on January 5, 2022

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