Proof of sum results

Question

I was going through some of my notes when I found both these sums with their results $$ x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1 $$ $$ 0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2} $$ I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not?

Answer 1

$\frac{1-x^{n+1}}{1-x}=1+x+x^2+\dots+x^n$, now if $n\to\infty$ and $|x|<1$ we get the first one.

Differentiate the first to get the second.

Answer 2

Hints:

$$(x^0+x^1+x^2+x^3+...x^n)(1-x)=x^0-x^1+x^1-x^2+x^2-x^3+x^3\cdots+x^{n}-x^{n+1}\\ =1-x^{n+1}$$ and $$(x^0+2x^1+3x^2+4x^3+...(n+1)x^n)(1-2x+x^2)\\ =x^0-2x^1+x^2+2x^1-4x^2+2x^3+3x^2-6x^3+3x^4\cdots\\+(n+1)x^n-2(n+1)x^{n+1}+(n+1)x^{n+2}\\ =1-(n+2)x^{n+1}+(n+1)x^{n+2}.$$

Answer 3

we have $\sum_{i=0}^nx^i=\frac{x^{n+1}-1}{x-1}$ and the limit for $n$ tends to infinity exists if $|x|<1$

Answer 4

Others have answered on how to justify that $S=1+x+x^2+\cdots+x^n+\cdots$ equals $\frac{1}{1-x}$ for $|x|<1$. To get the second identity without differentiation, note that \begin{align*} 1+2x+3x^2+4x^3+\cdots&=1+(x+x)+(x^2+x^2+x^2)+(x^3+x^3+x^3+x^3)+\cdots\\ &=(1+x+x^2+x^3+\cdots)+(x+x^2+x^3+\cdots)+(x^2+x^3+\cdots)+\cdots\\ &=S+xS+x^2S+\cdots=S(1+x+x^2+\cdots)\\ &=S^2\\ &=\frac{1}{(1-x)^2}\cdot \end{align*}

Answer 5

Let $S=1+x+x^2+\dotsb$. Now, take a look at this:

$\phantom xS=1+x+x^2+x^3+\dotsb$
$xS=\phantom{1+{}}x+x^2+x^3+\dotsb$

Thus, we have: \begin{align} S&=1+xS\\ S-Sx&=1\\ S(1-x)&=1\\ S&=\frac1{1-x} \end{align}

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Moving on to the second one: This one is a bit trickier, but we'll use the identity $(n+2)-2(n+1)+n=0$ in a clever way.

Let $T=x+2x^2+3x^3+\dotsb$. Now:

$\phantom{(1-2x+x^2)}T=x+2x^2+3x^3+4x^4+5x^5+\dotsb$
$\phantom{(1+x^2)}-2xT=\phantom{x}-2x^2-4x^3-6x^4-8x^4-\dotsb$
$\phantom{(1-2x+{})}x^2T=\phantom{x+2x^2+3}x^3+2x^4+3x^5+\dotsb$

Add them up:

$(1-2x+x^2)T=x$

Thus, we have: $$T=\frac x{1-2x+x^2}=\frac x{(1-x)^2}$$

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EDIT: By the way, we haven't been paying much attention to the matter of convergence. All we've proven, in fact, is that where $S$ and $T$ converge, their sums are equal to what we've derived above.

We've actually implicitly assumed that $\displaystyle\lim_{N\to\infty}x^N=0$, which is only true where $-1<x<1$. This happens to be where $S$ and $T$ converge. (We should check the boundary points — that is, $x=1$ and $x=-1$ — but it's not too hard to see that the sums diverge there.)

Answer 6

There are many excellent proofs given above.

Another approach would be to note that the summation (LHS) is actually a binomial expansion of the result (RHS). (NB - This would only be applicable if you have to prove that the result is true, instead of having to find the result.)

Hence

$$\begin{align} \color{blue}{\frac 1{1-x}}&=(1-x)^{-1}\\ &=\sum_{r=0}^{\infty}{-1\choose r}(-x)^r\\ &=\color{lightgray}{1+(-1)(-x)+\frac{-1\cdot-2}{1\cdot 2}(-x)^2+\frac{-1\cdot -2\cdot -3}{1\cdot 2\cdot 3}(-x)^3\cdots}\\ &=\sum_{r=0}^{\infty}(-1)^r{r\choose r}(-x)^r \qquad\text{using upper negation}\\ &=\color{green}{\sum_{r=0}^{\infty}{r\choose 0}x^r} =\sum_{r=0}^{\infty}x^r \\ &=1+x+x^2+x^3+\cdots+x^r+\cdots \\ &\phantom{=}\text{(Geometric Series)} \end{align}$$

and

$$\begin{align} \color{blue}{\frac 1{(1-x)^2}}&=(1-x)^{-2}\\ &=\sum_{r=0}^{\infty}{-2\choose r}(-x)^r\\ &=\color{lightgray}{1+(-2)(-x)+\frac{-2\cdot-3}{1\cdot 2}(-x)^2+ \frac{-2\cdot-3\cdot -4}{1\cdot 2\cdot 3}(-x)^3\cdots}\\ &=\sum_{r=0}^{\infty}(-1)^r{r+1\choose r}(-x)^r \qquad\text{using upper negation}\\ &=\color{green}{\sum_{r=0}^{\infty}{r+1\choose 1}x^r}=\sum_{r=0}^{\infty}(r+1)x^r \\ &=1+2x+3x^2+4x^3+\cdots+(r+1)x^r+\cdots \\ &\phantom{=}\text{(Arithmetico-Geometric Series)}\\ \end{align}$$

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Similarly,

$$\begin{align} \color{blue}{\frac 1{(1-x)^3}}&=(1-x)^{-3}\\ &=\sum_{r=0}^{\infty}{-3\choose r}(-x)^r\\ &=\color{lightgray}{1+(-3)(-x)+\frac{-3\cdot-4}{1\cdot 2}(-x)^2+ \frac{-3\cdot-4\cdot -5}{1\cdot 2\cdot 3}(-x)^3\cdots}\\ &=\sum_{r=0}^{\infty}(-1)^r{r+2\choose r}(-x)^r \qquad\text{using upper negation}\\ &=\color{green}{\sum_{r=0}^{\infty}{r+2\choose 2}x^r}=\sum_{r=0}^{\infty}\frac{(r+2)(r+1)}2x^r \\ &=1+3x+6x^2+10x^3+\cdots+\frac{(r+2)(r+1)}2 x^r+\cdots \\ \end{align}$$ This can also be derived by differentiating the result of the Arithmetico-Geometric Series.

By extension, $$\color{blue}{\frac 1{(1-x)^{n+1}}}=\color{green}{\sum_{r=0}^{\infty}{r+n\choose n}x^r}$$