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I'm currently learning the prove of sum of geometric series on Khan Academy.

I understand the behaviour of the function when $|r| > 1$, when $|r| < 1$, when $r = 0$ and when $r = -1$, but I am a bit confused by its behaviour when $r = 1$.

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The narrator said that when $r = 1$, the limit function is undefined because the denominator of the limit function would be $0$, and the behaviour of the limit function is UNDEFINED, which I do understand.

My confusion arises when I tried to substitute $r = 1$ into the original function for sum of geometric series,

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if $r = 1$, then every term would equal to a, and the sum of the geometric series would approach infinity, so its behaviour is DEFINED.

So when $r = 1$, behaviour of sum function is DEFINED, but behaviour of limit function is UNDEFINED, but sum function also equal to limit function?!

This is causing me so much confusion.

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  • $\begingroup$ "proof" ≠ "prove". $\endgroup$ – user21820 Mar 23 at 13:21
  • $\begingroup$ I would argue both the series and the rational function give the value $\infty$ at $r=1$. Compactification makes working with limits of rational functions so much cleaner. $\endgroup$ – Brevan Ellefsen Mar 23 at 16:01
  • $\begingroup$ @BrevanEllefsen The function $1/1-x$ has no unique value at $1,$ even after compactifying $\mathbf R.$ At best, it has both $+\infty$ and $-\infty$ as candidate values. $\endgroup$ – Allawonder Mar 23 at 16:18
  • $\begingroup$ @Allawonder we don't compactify to two infinities - we compactify to a single infinity. The one point compactification of $\mathbb R$ $\endgroup$ – Brevan Ellefsen Mar 23 at 16:19
  • $\begingroup$ @BrevanEllefsen Oh, I did not realise that you meant the analogue of the stereo map. I thought you meant extending the reals to include both $+$ and $-\infty.$ $\endgroup$ – Allawonder Mar 23 at 16:21
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In order to understand why $r=1$ is not allowed you have to look at the proof of the geometric series (I will neglect the constant $a$). We start with

$$S_n = 1 + r+ r^2+ ... + r^n$$ $$rS_n = r + r^2 + r^3 +... + r^{n+1}$$

Then we subtract both equations.

$$S_n ( 1-r) = 1 - r^{n+1}$$

Solving for $S_n$ requires that $r\neq 1$ or we would dived by $0$.

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  • $\begingroup$ +1 for showing why the result fails for $r=1$. $\endgroup$ – Paramanand Singh Mar 23 at 12:02
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What happens is that the equality$$\sum_{k=0}^nar^n=\frac{a-ar^{n+1}}{1-r}$$only holds when $r\neq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $\displaystyle\sum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$\sum_{k=0}^na1^n=\sum_{k=0}^na=(n+1)a.$$

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  • $\begingroup$ hi Jose, thanks for helping out. Do you mind if I ask, does the equality hold for r = -1? My guess is that it doesn't, but just want to confirm with you. $\endgroup$ – Thor Mar 23 at 11:50
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    $\begingroup$ My first equality holds for every $r\neq1$. $\endgroup$ – José Carlos Santos Mar 23 at 11:53
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It is a matter of terminology. You are right to say that "its behaviour is DEFINED", but this is an informal statement, meaning that we understand that the series diverges to infinity.

But when the limit of a series cannot be expressed by a real number, we say that it is undefined. You must accept this convention.

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The best way is to look at an actual geometric series with ratio of 1, such as

$2+2+2+2+2+2+2...$

Here, because each term is simply the previous term multiplied by 1, the series diverges, no limit can be found for obvious reasons.

Take the common ratio of $-1$

$(1)+(-1)+(1)+(-1)+(1)...$

Here, the value bounces between 0 and 1, so no limit can be found.

Note: A limit ONLY occurs if the partial sum, i.e. the sum up to n terms, tends to some number as n gets larger and larger.

As this is not the case if $|r|=1$, the function simply has no limit.

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The only problem is that you are considering that infinite is a number. If something is infinite, then it's undefined. So, when r = 1, it's not possible to calcule its sum, because the generated infinite value it's not defined.

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