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limit of geometric series sum when r = 1

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1

$begingroup$

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.

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

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.

calculus sequences-and-series limits summation

share|cite|improve this question

asked 43 mins ago

ThorThor

24617

$endgroup$

add a comment | 

1

$begingroup$

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.

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

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.

calculus sequences-and-series limits summation

share|cite|improve this question

asked 43 mins ago

ThorThor

24617

$endgroup$

add a comment | 

$begingroup$

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.

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

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.

calculus sequences-and-series limits summation

share|cite|improve this question

asked 43 mins ago

ThorThor

24617

$endgroup$

share|cite|improve this question

asked 43 mins ago

ThorThor

24617

share|cite|improve this question

asked 43 mins ago

ThorThor

24617

asked 43 mins ago

ThorThor

24617

asked 43 mins ago

ThorThor

24617

3 Answers
3

active

oldest

votes

3

$begingroup$

What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$

share|cite|improve this answer

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  My first equality holds for every $rneq1$.
  $endgroup$
  – José Carlos Santos
  34 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $rneq 1$ or we would dived by $0$.

share|cite|improve this answer

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for showing why the result fails for $r=1$.
  $endgroup$
  – Paramanand Singh
  24 mins ago
  

  
  
  
  

add a comment | 

0

$begingroup$

You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.

Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.

share|cite|improve this answer

answered 13 mins ago

AllawonderAllawonder

2,256616

$endgroup$

add a comment | 

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3

$begingroup$

What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$

share|cite|improve this answer

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  My first equality holds for every $rneq1$.
  $endgroup$
  – José Carlos Santos
  34 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$

share|cite|improve this answer

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  37 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  My first equality holds for every $rneq1$.
  $endgroup$
  – José Carlos Santos
  34 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

What happens is that the equality$$sum_{k=0}^nar^n=frac{a-ar^{n+1}}{1-r}$$only holds when $rneq1$. When $r=1$, it doesn't make sense. So, in order to study the behaviour of the series $displaystylesum_{k=0}^nar^n$ when $r=1$, we have to take another apprach. And that approach is:$$sum_{k=0}^na1^n=sum_{k=0}^na=(n+1)a.$$

share|cite|improve this answer

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

$endgroup$

share|cite|improve this answer

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

answered 39 mins ago

José Carlos SantosJosé Carlos Santos

169k23132237

2

$begingroup$

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 $rneq 1$ or we would dived by $0$.

share|cite|improve this answer

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for showing why the result fails for $r=1$.
  $endgroup$
  – Paramanand Singh
  24 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

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 $rneq 1$ or we would dived by $0$.

share|cite|improve this answer

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  +1 for showing why the result fails for $r=1$.
  $endgroup$
  – Paramanand Singh
  24 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

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 $rneq 1$ or we would dived by $0$.

share|cite|improve this answer

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

$endgroup$

share|cite|improve this answer

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

answered 30 mins ago

MachineLearnerMachineLearner

1,071110

0

$begingroup$

You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.

Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.

share|cite|improve this answer

answered 13 mins ago

AllawonderAllawonder

2,256616

$endgroup$

add a comment | 

0

$begingroup$

You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.

Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.

share|cite|improve this answer

answered 13 mins ago

AllawonderAllawonder

2,256616

$endgroup$

add a comment | 

$begingroup$

You're getting confused because you assume $infty$ is a real number. But it is not. The sum of a (real) series is useful only if it converges to a real number. Thus, since the function $$frac{1}{1-x}$$ is not a real number for $x=1,$ neither does it behave regularly near that point, it follows that you cannot use it to study the original series except for where the series converges to a real number.

Here you're learning the important lesson that even otherwise great men like Euler didn't quickly learn; namely that the "sum" of a series represents it only at those points where the series converges. Otherwise, the function represents something else.

share|cite|improve this answer

answered 13 mins ago

AllawonderAllawonder

2,256616

$endgroup$

share|cite|improve this answer

answered 13 mins ago

AllawonderAllawonder

2,256616

answered 13 mins ago

AllawonderAllawonder

2,256616

answered 13 mins ago

AllawonderAllawonder

2,256616