Proof of the power series 1 + $x^2$ + $x^3$ + $\ldots$ + $x^n$ = $\frac{1}{1-x}$

Question

Can anyone show me the proof of this equation:

$$ \lim_{n \to \infty} 1 + x + x^2 + x^3 + \ldots + x^n = \frac{1}{1-x}, $$

where $|x|<1$.

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Edit: I have then additionally written $x$ on the left side.

Answer 1

I am not sure if it is just a notational error, or a genuine misunderstanding, but the correct formula is $$1 + x + x^2 + x^3 + \cdots + x^n + \cdots = \frac{1}{1-x}$$ Notice that unlike the question in the OP, the sum on the left does not stop at the $x^n$ term, but continues.

As to why this is true, there are a few different approaches of varying levels of formality. One (rather nonrigorous) approach is to start by calling the expression on the left $S$, and thinking about what happens if you multiply $S$ by $x$. We have $$S = 1 + x + x^2 + x^3 + \cdots$$ $$xS = x + x^2 + x^3 + \cdots$$

Subtracting the bottom equation from the top, we see that everything on the right cancels except for the $1$ in the top equation; that is, we have $$S - xS = 1$$ $$S(1-x) = 1$$ $$S = \frac{1}{1-x}$$ The reason I say this is a rather nonrigorous approach is that it begins by assuming that the infinite sum of the left-hand side converges to some number $S$. But how do we know it converges at all? In fact, it doesn't converge if $|S| \ge 1$, and in that case this entire argument is bogus. Another problem with this approach is that it rather casually assumes that you can just cancel the sums of one infinite series off of the sums of the other; again, one really needs to give careful consideration to issues of convergence in order to justify this rigorously.

A second, slightly different way of approaching this is to consider the expression $$(1-x)(1 + x + x^2 + x^3 + \cdots)$$ Using the distributive property one gets $$(1 + x + x^2 + x^3 + \cdots) - (x + x^2 + x^3 + \cdots)$$ and again everything cancels except the $1$ in the first pair of parentheses, so $$(1-x)(1 + x + x^2 + x^3 + \cdots) = 1$$ from which the desired conclusion follows.

This second approach is really just a slightly rewritten version of the first approach, and suffers from the same problems.

A more careful approach needs to begin with the formula for a finite series. Call $S_n$ the sum of the first $n$ terms. Then

$$S_n = 1 + x + x^2 + x^3 + \cdots + x^n = \frac{1-x^{n+1}}{1-x}$$ which can be proven using a variation of the two approaches described above. Then we define $1 + x + x^2 + \cdots$ to mean $\lim_{n\to \infty} S_n$. If $|x|<1$ then $\lim_{n\to\infty} x^{n+1} = 0$, so $$\lim_{n\to\infty}S_n = \lim_{n\to\infty} \frac{1-x^{n+1}}{1-x} = \frac{1}{1-x}$$

Answer 2

This is not true unless $x=0$. You have only a finite sum on the left, which is not a "series".

But we do have $$ \sum_{n=0}^\infty x^n=\frac{1}{1-x} $$ for $|x|<1$.

The keyword is Geometric series. And the detailed proof is given by the Wikipedia article.

Answer 3

I'm assuming you missed the term x in your question, otherwise the final form will be slightly different, you could just subtract an x from it

$1 + x + x^2 + x^3 + \ldots + x^n = $

$[1 + x + x^2 + x^3 + \ldots + x^n]\frac{1-x}{1-x} = $

$[1 - x + x - x^2 + x^2 - x^3 + x^3 + \ldots - x^n + x^n - x^{n+1}]\frac{1}{1-x} = $

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

for |x| < 1 and n tends to infinity, this becomes $\frac{1}{1 - x}$

Answer 4

First, there's a missing $x$ term in your sum - it should be $1+x+x^2+x^3+\cdots+x^n$. Secondly, the left hand side is then equal to $\frac{1-x^{n+1}}{1-x}$, as can be seen by multiplying both sides by $1-x$: \begin{align*} 1-x^{n+1}&=(1-x)(1+x+x^2+\cdots+x^n) \\ &=1+x+x^2+\cdots+x^n-(x+x^2+\cdots+x^{n+1}) \\ &=1-x^{n+1}, \end{align*} verifying the identity, a specific case of the difference of n-th powers: $$x^n-y^n=(x-y)\sum\limits_{k=0}^{n-1}x^ky^{n-1-k}.$$

Note, if you've done some calculus, if we take the limit as $n\to\infty$, \begin{align*}\lim_{n\to\infty} 1-x^{n+1} &= \lim_{n\to\infty} (1-x)(1+x+\cdots+x^n) \\ &= (1-x)\lim_{n\to\infty}(1+x+\cdots+x^n) \\ &= (1-x)\sum_{k=0}^\infty x^k, \end{align*} and since the left hand side above, for $\lvert x\rvert<1$, goes to 1 as $n\to\infty$ (take a number between -1 and 1 and raise it to a very high power and it goes to 0), we have the formula for the geometric series, $$\frac{1}{1-x}=\sum_{k=0}^\infty x^k.$$

But you have to go "all the way to infinity" for equality to hold.