Partial Sums of Geometric Series

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 $$a^{n}-1=(a-1)(a^{n-1} + a^{n-2}+ cdots + 1) Rightarrow a^{n-1} + a^{n-2}+ cdots + 1 = frac{a^{n}-1}{a-1}$$

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.

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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.

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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.

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

⇒ $(1-x)S_n(x)=1-x^{n+1}$

⇒ $S_n(x)=frac{1-x^{n+1}}{1-x}$

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,

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

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.