I A question regarding the ratio test for limits

[MathematicalPhysicist]

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.

 

[member 587159]

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]

@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.

 

[Mark44]

@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]

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.

 

[member 587159]

@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.