My Math Forum radius of convergence with indeterminate form

 Calculus Calculus Math Forum

 January 10th, 2017, 08:05 AM #1 Member   Joined: Jan 2017 From: California Posts: 62 Thanks: 5 radius of convergence with indeterminate form 20170110_080156.jpg The limit of this ratio test for this series suggests that for any x except zero the series diverge. The thing I cannot get over is the following: What if I choose an x that is so infinitely close to zero such as it dominates infinity? Such as that the product of the two factors is less than 1. I believe this was the case of an indeterminate form when L'Hôpital's rule was necessary in limits calculations. Thanks in advance. Last edited by skipjack; January 10th, 2017 at 08:46 AM.
January 10th, 2017, 08:57 AM   #2
Math Team

Joined: Jan 2015
From: Alabama

Posts: 2,487
Thanks: 630

Quote:
 Originally Posted by dthiaw Attachment 8341 The limit of this ratio test for this series suggests that for any x except zero the series diverge. The thing I cannot get over is the following: What if I choose an x that is so infinitely close to zero such as it dominates infinity?
Where did you get the idea that such a number exists? x is a fixed real number. It sounds like you are talking about x being an "infinitesimal" or you are taking a limit. Neither of those is true for x being a fixed real number.

Quote:
 Such as that the product of the two factors is less than 1. I believe this was the case of an indeterminate form when L'Hôpital's rule was necessary in limits calculations. Thanks in advance.

 January 10th, 2017, 09:02 AM #3 Member   Joined: Jan 2017 From: California Posts: 62 Thanks: 5 What if I choose an x so small such as 0.00000.... with the 1 all the way to the north pole? Can I do that? I assume I could since the value of x could be continuous and not discrete. Couldn't that small number overpower infinity (0 vs infinity) type of fight? Last edited by skipjack; January 10th, 2017 at 10:50 AM.
 January 10th, 2017, 10:01 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra Two approaches to put you on the right trackWhat can you say about the sequence $a_n=cn$ where $c \ne 0$? In particular, what can you say about the expression $a_{k+1}-a_k$? How many zeroes does it take to get to the 1 to the North Pole? A million? A billion? A googol? Let's say that the answer is $m$ zeros. $m$ is clearly a finite number. And $0.000\ldots01$, with $m$ zeroes (including the one to the left of the decimal point) is equal to $\frac1{10^m}$. What is the limit $$\lim_{n \to \infty} \frac{n}{10^m}$$ Alternatively, what does it mean in terms of the definition of the limit for $$\lim_{n \to \infty} \frac{n}{10^m}=1$$ Thanks from dthiaw
January 10th, 2017, 10:25 AM   #5
Math Team

Joined: Jan 2015
From: Alabama

Posts: 2,487
Thanks: 630

Quote:
 Originally Posted by dthiaw What if I choose an x so small such as 0.00000.... with the 1 all the way to the north pole? Can I do that? I assume I could since the value of x could be continuous and not discrete. Couldn't that small number overpower infinity (0 vs infinity) type of fight?
Can you quantify the number of "0"s required to go "all the way to the north pole"? You can say a very large, but specific, number of "0"s. However, that would not "overpower" infinity. Any number "times infinity" is "infinity". I put "times infinity" and "infinity" in quotes because "infinity" is NOT a real number and multiplication of any real number with "infinity" is not defined.

Last edited by skipjack; January 10th, 2017 at 10:52 AM.

January 10th, 2017, 11:21 AM   #6
Member

Joined: Jan 2017
From: California

Posts: 62
Thanks: 5

Quote:
 Originally Posted by Country Boy Can you quantify the number of "0"s required to go "all the way to the north pole"? You can say a very large, but specific, number of "0"s. However, that would not "overpower" infinity. Any number "times infinity" is "infinity". I put "times infinity" and "infinity" in quotes because "infinity" is NOT a real number and multiplication of any real number with "infinity" is not defined.
Quote:
 Originally Posted by v8archie Two approaches to put you on the right trackWhat can you say about the sequence $a_n=cn$ where $c \ne 0$? In particular, what can you say about the expression $a_{k+1}-a_k$? How many zeros does it take to get to the 1 to the North Pole? A million? A billion? A googol? Let's say that the answer is $m$ zeros. $m$ is clearly a finite number. And $0.000\ldots01$, with $m$ zeros (including the one to the left of the decimal point) is equal to $\frac1{10^m}$. What is the limit $$\lim_{n \to \infty} \frac{n}{10^m}$$ Alternatively, what does it mean in terms of the definition of the limit for $$\lim_{n \to \infty} \frac{n}{10^m}=1$$
Ok I understand what you guys are saying. (that any finite number no matter how small it is multiplied by infinity = infinity)

However, there was an indeterminate case that forced the use of L'Hôpital's in limits calculation. It was stating that 0+ multiplied by infinity is not equal to infinity and that's why we had to rearrange things and present them in the form 0/0 or infinity over infinity. I feel like this is contradicting the stuff we're discussing.

Last edited by skipjack; January 10th, 2017 at 11:29 AM.

 January 10th, 2017, 12:06 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,778 Thanks: 2195 Math Focus: Mainly analysis and algebra If $f(x) \to 0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x)$$ is the indeterminate case $0 \cdot \infty$. This is the case you are thinking of. But that is because the limit of $f(x)$ is equal to zero, not because $f$ is very small. But if $f(x)=c \ne 0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x)$$ is the undefined (but not indeterminate) case $c \cdot \infty = \pm\infty$. And if $f(x)=0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} 0 \cdot g(x) = \lim_{x \to a} 0 = 0$$ Thanks from dthiaw
January 10th, 2017, 12:28 PM   #8
Member

Joined: Jan 2017
From: California

Posts: 62
Thanks: 5

Quote:
 Originally Posted by v8archie If $f(x) \to 0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x)$$ is the indeterminate case $0 \cdot \infty$. This is the case you are thinking of. But that is because the limit of $f(x)$ is equal to zero, not because $f$ is very small. But if $f(x)=c \ne 0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x)$$ is the undefined (but not indeterminate) case $c \cdot \infty = \pm\infty$. And if $f(x)=0$ and $g(x) \to \infty$ then $$\lim_{x \to a} f(x)g(x) = \lim_{x \to a} 0 \cdot g(x) = \lim_{x \to a} 0 = 0$$
Thank you so much V8archie. it s finally clear. You are the man

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post MrDrProfessorPatrick Calculus 7 May 3rd, 2016 01:45 PM MrDrProfessorPatrick Calculus 1 March 8th, 2016 04:50 AM MustafaMotani Calculus 2 October 31st, 2014 09:35 AM GeniusBoy Calculus 20 October 5th, 2014 07:05 PM rain Calculus 14 August 1st, 2013 12:45 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top