I A question regarding the ratio test for limits

MathematicalPhysicist

Gold Member
4,032
121
So we have the theorem:
if ##a_n>0## and ##\lim_{n\to \infty} a_{n+1}/a_n = L## then ##\lim_{n\to \infty} a_n^{1/n}=L##.

Now, the proof that I had seen for ##L\ne0## that we choose ##\epsilon<L##.

But what about the case of ##\epsilon>L##, in which case we have:
##a_{n+1}>(L-\epsilon)a_n## but the last RHS is negative, so I cannot take the n-th root without going into problems of an n-th root of a negative number, which is not defined for even n's in the real line.

I read this solution from Albert Blank's solutions to Fritz John and Richard Courant's textbook.
 

Math_QED

Homework Helper
1,020
323
So we have the theorem:
if ##a_n>0## and ##\lim_{n\to \infty} a_{n+1}/a_n = L## then ##\lim_{n\to \infty} a_n^{1/n}=L##.

Now, the proof that I had seen for ##L\ne0## that we choose ##\epsilon<L##.

But what about the case of ##\epsilon>L##, in which case we have:
##a_{n+1}>(L-\epsilon)a_n## but the last RHS is negative, so I cannot take the n-th root without going into problems of an n-th root of a negative number, which is not defined for even n's in the real line.

I read this solution from Albert Blank's solutions to Fritz John and Richard Courant's textbook.
If ##L>0##, you can always choose ##\epsilon>0## such that ##0<\epsilon <L##. What exactly is the problem here?
 

MathematicalPhysicist

Gold Member
4,032
121
@Math_QED the definition of a limit is ##\lim_{n\to \infty}b_n=L \Leftrightarrow \forall \epsilon >0 \exists N(\epsilon) \in \mathbb{N} (n>N(\epsilon) \rightarrow |b_n-L|<\epsilon##.

Then for the definition of limit I need to show that limit also applies for ##\epsilon>L##, since the definition requires that the statement ##n>N(\epsilon)\rightarrow |b_n-L|<\epsilon## will be true for every positive epsilons not only those that are less than ##L##.
And I don't see why does this follow here?

Perhaps I am confused.
 
32,108
3,993
@Math_QED the definition of a limit is ##\lim_{n\to \infty}b_n=L \Leftrightarrow \forall \epsilon >0 \exists N(\epsilon) \in \mathbb{N} (n>N(\epsilon) \rightarrow |b_n-L|<\epsilon##.

Then for the definition of limit I need to show that limit also applies for ##\epsilon>L##, since the definition requires that the statement ##n>N(\epsilon)\rightarrow |b_n-L|<\epsilon## will be true for every positive epsilons not only those that are less than ##L##.
And I don't see why does this follow here?

Perhaps I am confused.
Maybe so. The definition says, in part, "for any positive ##\epsilon##", but you want to show that for reasonably large n, that ##b_n## and L are only a small distance apart. The concept here is that no matter how close together someone else requires these two numbers to be, you can find a number n that forces ##b_n## and L to be that close. There is no reason for someone to choose ##\epsilon## to be large; i.e., larger than L.

Here's an example. Let ##b_n = \frac 1 2, \frac 2 3, \frac 3 4, \dots, \frac n {n + 1}, \dots##. The limit of this sequence clearly is 1. If someone else chooses ##\epsilon = 2##, how far along in the seqence do you need to go so that ##|b_n - 1| < 2##? If they want to make you work, they will choose a much smaller value for ##\epsilon##.
 

MathematicalPhysicist

Gold Member
4,032
121
Maybe so. The definition says, in part, "for any positive ##\epsilon##", but you want to show that for reasonably large n, that ##b_n## and L are only a small distance apart. The concept here is that no matter how close together someone else requires these two numbers to be, you can find a number n that forces ##b_n## and L to be that close. There is no reason for someone to choose ##\epsilon## to be large; i.e., larger than L.

Here's an example. Let ##b_n = \frac 1 2, \frac 2 3, \frac 3 4, \dots, \frac n {n + 1}, \dots##. The limit of this sequence clearly is 1. If someone else chooses ##\epsilon = 2##, how far along in the seqence do you need to go so that ##|b_n - 1| < 2##? If they want to make you work, they will choose a much smaller value for ##\epsilon##.
I need to relearn stuff that I have forgotten.
 

Math_QED

Homework Helper
1,020
323
@Math_QED the definition of a limit is ##\lim_{n\to \infty}b_n=L \Leftrightarrow \forall \epsilon >0 \exists N(\epsilon) \in \mathbb{N} (n>N(\epsilon) \rightarrow |b_n-L|<\epsilon##.

Then for the definition of limit I need to show that limit also applies for ##\epsilon>L##, since the definition requires that the statement ##n>N(\epsilon)\rightarrow |b_n-L|<\epsilon## will be true for every positive epsilons not only those that are less than ##L##.
And I don't see why does this follow here?

Perhaps I am confused.
Show that this definition is equivalent with the definition:

##\forall \epsilon \in (0,k): \exists N: \dots##

where ##k>0## is some fixed constant.
 

Want to reply to this thread?

"A question regarding the ratio test for limits" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top