Proof of $\frac{1}{(1-r)^2}$

Question

I was given that $S=1+2r+3r^2+\cdots = \frac{1}{(1-r)^2}$ was an identity from my students, and I tried to prove directly using geometric series.

I got stuck and looked around online only to be told that taking the derivative of the series that converges to $\frac{1}{1-r}$.

I liked that proof because it was simple and easy, but unfortunately my student is in Alg II so I cannot explain her with derivatives.

My idea was to expand $S$ as $$(1+r+r^2+r^3+\cdots) +(r+r^2+r^3+\cdots)+(r^2+r^3+\cdots)+\cdots$$

but I get confused whether I should rewrite this as $$(1+r+r^2+r^3+\cdots) +r(1+r+r^2+\cdots)+r^2(1+r+\cdots)+\cdots$$

or

$$(1+r+r^2+r^3+\cdots) +((-1)+1+r+r^2+\cdots)+((-1-r)+1+r+\cdots)+\cdots$$

I was able to get the ID using the first way, but I am not sure if the step I took was valid due to some stuff I vaguely remember about manipulating an infinite series like that was not always allowed.

And on the second way I simply did not know where to go from there.

Can I have some help ?

It would also be helpful to know the criteria of when I can manipulate the order of how to add the series.

Answer 1

Your first approach $$1(1+r+r^2+\cdots)+r(1+r+r^2+\cdots)+r^2(1+r+r^2+\cdots)+\cdots$$ is the best one for intuition. Pulling out the common factor then gives us $$(1+r+r^2+\cdots)(1+r+r^2+\cdots)=(1+r+r^2+\cdots)^2,$$ at which point we can use the standard geometric series result for $|r|<1$.

This is perfectly fine to do, since we're dealing with absolutely convergent series, which is equivalent to unconditionally convergent (meaning we can do such rearrangements without fear of changing the result).

Answer 2

There is some good advice in the comments. As for rearrangements and absolute convergence, the key insight is this: For a a family $(x_i)_{i\in I}$ of non-negative numbers indexed by any set $I$, you can define the sum: $$\sum_{i\in I} x_i=\sup_{F\Subset I}\sum_{i\in F}x_i,$$ where $F\Subset I$ indicates that $F$ is a finite subset of $I$. This definition is clearly independent of any order, and it is easy to see that for a sequence of nonnegative numbers, one gets $$\sum_{i\in\mathbb{N}} x_i=\sum_{i=1}^\infty x_i.$$ And that is why the sums of nonnegative sequences do not depend on summation order. For absolutely convergent sequences, you apply this to the positive and negative parts separately, and get the same result. Moreover, this principle will also apply to double sums as well, so your treatment of the sum as an infinite sum of sums of infinite series is completely okay.

Answer 3

You can manipulate the series in this case if you prove that it's absolutely convergent first (which it is, for appropriate values of $r$!) However, I suppose you aren't dealing in this level of mathematics with this student, than I suppose you should only suppose it's valid and go forward with it. I think a much clearer notation for "breaking up the sum" that should be quite enjoyed by the students is thinking about it in this triangle form:

$1$

$r \space \space r$

$r^2 \space \space r^2 \space \space r^2$

$r^3 \space \space r^3 \space \space r^3 \space \space r^3 \space \space$

...

Just sum each column like a simple geometric series, than sum those results to get the final answer. It's exactly the same thing you are doing, but I think it is easier to see that both ways of summing are the same thing.

I would then proceed to "prove" that 0 = 1, by manipulating a non absolute convergent series, like:

$(1-1) + (1-1) + (1-1) + ... = 0$

$1 - (1 - 1+1- 1+1-1+...) = 0$

$1 - [(1 - 1)+(1- 1)+(1-1)+...] = 1 - 0=1 = 0$

I'm not a teacher, hence I'm not very pedagogical, but I think you should keep them on their toes about messing up with infinite series, as "it's not always okay to do so"

Answer 4

Formerly:

$$\sum_{n=1}^{\infty}nr^{n-1}=\sum_{n=1}^{\infty}\sum_{k=1}^{n}r^{n-1}=\sum_{k=1}^{\infty}\sum_{n=k}^{\infty}r^{n-1}=\sum_{k=1}^{\infty}r^{k-1}\sum_{n=1}^{\infty}r^{n-1}=\left(\sum_{n=1}^{\infty}r^{n-1}\right)\left(\sum_{k=1}^{\infty}r^{k-1}\right)=\left(1-r\right)^{-2}$$

It is based on your first approach.

addendum:

To explain it to $15$ year oldies I should say that your first approach works fine:

Denote $s=\sum_{n=1}^{\infty}r^{n-1}$ and keep in mind that $s=(1-r)^{-1}$

Then $\sum_{n=1}^{\infty}nr^{n-1}=1.s+r.s+r^{2}.s+...=s\left(1+r+r^{2}+...\right)=s.s=s^{2}=(1-r)^{-2}$

Answer 5

this is an arithmetico geometric progression of the form $$a+(a+d)r+(a+2d)r^2+....\infty$$ where the sum is given by $$S_{\infty}=\frac{a}{1-r}+\frac{dr}{(1-r)^2}$$ to prove this let $S_n$ denote the sum of the first $n$ terms of the series then $$S_n=a+(a+d)r+(a+2d)r^2+.......[a+(n-1)d]r^{n-1}$$ $$r S_n=ar+(a+d)r^2+(a+2d)r^3+.......[a+(n-1)d]r^{n}$$ subtracting both $$(1-r) S_n=a+dr+dr^2+.......dr^{n-1}-[a+(n-1)d]r^n $$ $$=a+\frac{dr(1-r^{n-1})}{1-r}-[a+(n-1)d]r^n$$ or $$S_n= \frac{a}{1-r}+\frac{dr}{(1-r)^2} - \frac{dr^n}{(1-r)^2}-\frac{[a+(n-1)d]r^n}{(1-r)^2}$$ now when $r<1$ and $n \to \infty$ you get the required.

Answer 6

For $\lvert r\rvert< 1$, \begin{align*} (1-r)^2\sum_{n=0}^\infty (n+1) r^n &= \sum_{n=0}^\infty (n+1)r^n - 2r \sum_{n=0}^\infty (n+1)r^n + r^2 \sum_{n=0}^\infty (n+1)r^n \\ &= \sum_{n=0}^\infty (n+1)r^n - 2 \sum_{n=1}^\infty n r^n + \sum_{n=2}^\infty (n-1)r^n \\ &= 1+2r-2r+\sum_{n=2}^\infty \underbrace{\left( (n+1)r^n - 2nr^n + (n-1)r^n \right)}_{=0} \\ &= 1 \end{align*}

Answer 7

Notice that $$(1+2r+3r^2+4r^3+\dots)\cdot(1-r)=1+(2-1)r+(3-2)r^2+(4-3)r^3+\cdots=\\ =1+r+r^2+r^3+\cdots=\frac{1}{1-r}\,,$$ which immediatly implies the required identity $$ 1+2r+3r^2+4r^3+\cdots=\frac{1}{(1-r)^2}\,.$$