# Why is the sum of an infinite geometric series of fractions = zero?

#### B9766

I'm having trouble getting my mind around a concept in an introductory lesson in pre-calculus.

The professor says that $\displaystyle \sum_{i=0}^{\infty}r^i = \dfrac{1}{1-r}\ \$for$\ \ |r| < 1$

He goes on to prove this with $1 +r+r^2+r^3...+r^{n} = x\ \$ (eq 1)

Then, multiplying by r: $r+r^2+r^3...+r^{n+1} = rx\ \$ (eq 2)

Subtracting (eq 1) from (eq 2): $1 + r^{n+1} = rx - x\ \$ (eq 3)

Dividing both sides by $r-1$: $x=\dfrac{1+ r^{n+1}}{r-1}\ \$(eq 4)

And switching signs: $x=\dfrac{1- r^{n+1}}{1-r}\ \$ (eq 5)

All good. But then he says, "since the absolute value of $r$ is less than 1, the value of $r^{n+1}$ at infinity is zero." (!???)

"And therefore, $x=\dfrac{1}{1-r}\ \$" (eq 6)

This makes no sense to me. It seems to me that because a variable approaches zero, it does not mean the variable is ever equal to zero. It may be infinitesimally small but it's still not zero.

He goes on to prove, using the same logic, that 0.999999999...$\infty=1$

Why wouldn't any number, no matter how small, still be greater than zero?

The professor's logic would seem to dictate that $1/\infty=0\ \$ which, in turn would imply $1/0 = \infty\ \$ and that flies in the face of logic and mathematics.

Perhaps it requires an understanding of calculus to comprehend this. I would appreciate any clarification.

#### HallsofIvy

MHF Helper
The problem is the phrase "the value of $$\displaystyle r^{n+1}$$ at infinity is zero.".

I personally would not say it that way because "infinity" is not actually a real number and you cannot calculate or say anything about the value of a function of a real variable "at infinity". The phrase "at infinity" is being used, here, as a short way of saying "the limit as n goes to infinity". Any number, with absolute value less than 1, taken to higher powers, gets smaller and smaller getting, for sufficiently large n, arbitrarily close to 0. The limit is 0.

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#### B9766

I agree. But without substituting zero for $r^{n+1}\ \$, the equation of $x = \dfrac{1}{1-r}\ \$ makes no sense.

#### Archie

The notation $\displaystyle \sum_{i=0}^{\infty}$ is defined to mean $\displaystyle \lim_{n \to \infty}\sum_{i=0}^{n}$, so the distinction you are drawing is a red herring.

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#### Plato

MHF Helper
I'm having trouble getting my mind around a concept in an introductory lesson in pre-calculus.
The professor says that $\displaystyle \sum_{i=0}^{\infty}r^i = \dfrac{1}{1-r}\ \$for$\ \ |r| < 1$
I agree. But without substituting zero for $r^{n+1}\ \$, the equation of $x = \dfrac{1}{1-r}\ \$ makes no sense.
I agree that none of this makes any sense if you are not grounded is basic concepts.
The series ${S_N} = \sum\limits_{k = 1}^N {{a_k}}$ has sum (is convergent) if and only if $\mathop {\lim }\limits_{N \to \infty } {S_N} = L < \infty$
If you are only in preCalculus without the benefit of understanding limits then there is no reason for you to be asked to understand this material.
But say you are. Then the next step is: for all $|r|<1$ we have as fact $\Large\mathop {\lim }\limits_{N \to \infty } {r^N} = 0$ That is the most important fact in all of this.

Here goes:
$\begin{array}{l}{S_N} = \sum\limits_{k = 1}^N {{r^k}} = r + {r^2} + {r^3} \ldots {r^N}\\r{S_N} = {r^2} + {r^3} + \cdots {r^N} + {r^{N + 1}}\\{S_N} - r{S_N} = r - {r^{N + 1}}\\{S_N} = \dfrac{{r - {r^N}}}{{1 - r}}\end{array}$

Now what happens as $N\to \infty$? Does $\mathop {\lim }\limits_{N \to \infty } {S_N} = \frac{r}{{1 - r}}~?$

This is usually given as if $|r|<1$ then $\sum\limits_{k = J}^\infty {A{r^k}} = \dfrac{{A{r^j}}}{{1 - r}}$
The sum is equal to the first term divided by one minus the common ratio.

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#### B9766

Thanks Plato.

As always, you show me where I misunderstood and you provide helpful guidance. These video courses aren't easy when it's impossible to ask the professor questions.

This particular lesson is titled "Sequences and Series" and introduces Limits. The chapter does not go into "converging" nor "diverging" limits. It does describe sigma, its notation and what it means in general terms. It is not much more enlightening than the discussion of asymptotes in the conics chapter. But it does describe limits in the context of infinite series.

The distinction you make does make sense to me.

I wish I could post the part of the video in question but it's proprietary.
In the part in question, the Professor shows, on the screen:

$\displaystyle x=\dfrac{1−r^{n+1}}{1−r} \ \ r<|1|\ \ \sum_{i=0}^{\infty}r^i=1+r+r^2+r^3+...=\dfrac{1}{1-r}$

And, verbatim, he says, "Now, imagine the following: Imagine the r is less than one and greater than minus one. In other words, the absolute value of r is one. That means that r to a large power starts getting smaller and smaller, doesn't it? Like, if r were a half, then one-half to a billionth power is really small. So, in the limit [emphasis mine], as we go to infinity, that r to the n plus 1 begins to disappear. And, in fact, if you add, from i equals zero to infinity, r to the i; that's one, plus r, plus r squared, plus r cubed, forever, the r to the n plus one disappears and your formula is one over one plus r. And, of course, you have to make sure the absolute value of r is less than one. If r is greater than one or equal to one in absolute value, the series doesn't add up to a finite number [emphasis mine]."

He goes on the describe Xeno's Paradox (with which I was already familiar) and concludes his description with, "...he argued you would never get across the room.And you can sort of see this as an infinite series."

The series appears on the screen: $\displaystyle \sum_{i=0}^{\infty}(\dfrac{1}{2})^i = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +... =?$

And he continues, "We're adding up one as the first step, and then a half, and then a quarter, and then an eighth, et cetera. But we know better now. Don't we? Now that we've studied these infinite series, we're adding this up to infinity. And, of course, if you add up to infinity, it's one over two, one-half to the i, from i equals zero to infinity. According to our formula, it's one over one minus r. Which is one over one minus one-half. And that equals 2. So, this infinite series, out to infinity, adds up to two. Xeno does cross that two-meter walk."

But in truth, if I understand you correctly, Xeno is still correct. The closest you can approach that limit of 2 meters is just up to, but never quite at the 2-meter limit.

The professor uses the same logic to prove that 0.9999999999999999...$\infty\ \ =\ \ 1$ And he adds, "Wow, the sum of nine-tenths plus nine-one-hundredths, and so on to infinty, is equal to 1."

To my mind, this last statement is false. The limit of this decimal, consisting of an infinite number of 9's, does approach 1, but is never exactly equal to it.

Should it be assumed that any summation of an infinite number of items is always the limit that the sum approaches, even though it doesn't say "limit"? Is that the part I'm missing?

The significance of limits and their calculation isn't lost on me. But I need to make sure I comprehend the concept.

This is really long but I appreciate all your help.

#### Plato

MHF Helper
$\displaystyle x=\dfrac{1−r^{n+1}}{1−r} \ \ r<|1|\ \ \sum_{i=0}^{\infty}r^i=1+r+r^2+r^3+...=\dfrac{1}{1-r}$

The series appears on the screen: $\displaystyle \sum_{i=0}^{\infty}(\dfrac{1}{2})^i = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +... =?$

The professor uses the same logic to prove that 0.9999999999999999...$\infty\ \ =\ \ 1$ And he adds, "Wow, the sum of nine-tenths plus nine-one-hundredths, and so on to infinty, is equal to 1."

To my mind, this last statement is false. The limit of this decimal, consisting of an infinite number of 9's, does approach 1, but is never exactly equal to it.
In the sum $\displaystyle \sum_{i=0}^{\infty}(\dfrac{1}{2})^i = 1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} +... =?$
$A=1$ that is the first term is $1$ and the ratio is $\dfrac{1}{2}$ so $S=\dfrac{1}{1-\frac{1}{2}}=\dfrac{2}{2-1}=2$

If you see $0.9999999\cdots=9\cdot10^{-1}+9\cdot10^{-2}+9\cdot10^{-3}+9\cdot10^{-4}+9\cdot10^{-5}+\cdots$
The first term is $9\cdot10^{-1}$ ans the ratio is $10^{-1}$ so the $\text{SUM}=\dfrac{9\cdot10^{-1}}{1-10^{-1}}=\dfrac{9}{10-1}=1$

#### Archie

To my mind, this last statement is false. The limit of this decimal, consisting of an infinite number of 9's, does approach 1, but is never exactly equal to it.
I do like this argument, although at the same time it is the most infuriating.

What you probably haven't been told is that the Real numbers are defined as the set of all limits of convergent sequences of rational numbers. 0.999999... is not a number because it's a decimal, it is a number who's decimal form represents the infinite series $\displaystyle \sum_{n=1}{\infty} \frac{9}{10^n}$. Thus, the fact that the series converges to 1 guarantees that 0.999... = 1 litereally by definition.

Confusion on this point is the result of the fact that we are taught about real numbers and decimals from an early age without explanation of what they really are (with good reason - it's way too complicated for 6 year olds). So we grow up believing that the reals are the result of decimal representations rather than the converse, that decimal representations depict a method of producing the real number they represent.

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#### B9766

It's even more disappointing in that this is the second time I'm doing this. 50 years ago I took mathematics up through Calculus 2 at university. 3 years ago I decided I didn't really understand anything I learned back then. Corporate management duties, even in the software industry, don't call on using Lagrange transformations all that much. My priorities in school were more about learning how to do problems over why they are done that way.

Over the past 3 years, I've re-taken Algebra I $II, Geometry, Trigonometry and now Pre-Calculus with the focus on why things are done the way they are, and where and how the ideas originated. It was a Eureka moment for me to discover how Pythagoras actually figured out his eponymous formula. I don't learn formulas anymore without working through their derivations. It became very time-consuming to derive some of the Trig identities. I find I can't move forward in my studies if I don't thoroughly understand why each step is done as it is. Seeming inconsistencies are major stumbling blocks for me. All this is to say thank you. Clearly, 14 years of mathematics, plus another 3 of review, didn't cover this topic in enough depth. I'll dig deeper into real numbers. If you have any recommendations for me, I would appreciate it. #### greg1313$\displaystyle x=\dfrac{1−r^{n+1}}{1−r} \ \ r<|1|$... B9766, you intended to type$ x=\dfrac{1−r^{n+1}}{1−r} \ \ \ for \ \ \ |r| <1 \$ ...

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