MHB Proof that lim loga_n/n = 0 in epsilon delta language

[arsenaler]

Let $\,a>0\,,\,a\neq1\,$ be a real number. We can prove by using the continuity of $\ln n$ function that $\;\lim\limits_{n\to\infty}\dfrac{\log_an}n=0\;$

However, this problem appears in my problems book quite early right after the definition of $\epsilon$-language definition of limit of a sequence, the reader is supposed not to know anything about continuity.

My question is: Is there any proof for this result in $\epsilon-delta$ language that is more elementary?

Please help me.

Thanks.

 

[HOI]

There won't be any "$\delta$" because the limit is being taken as n goes to infinity, not any finite value. What you want instead is to show that, given $\epsilon> 0$ there exist N such that if n>N then $|\frac{ln_a(n)}{n}|< \epsilon$.

Further, the use of "n" rather than "x" implies that this a sequence, not a function of x.

 

[HOI]

Since n and $log_a(n)$, for n> 1, are positive we can write that as $\frac{log_a(n)}{n}< \epsilon$ and then $log_a(n)< n\epsilon$.

 

[Prove It]

Let $\,a>0\,,\,a\neq1\,$ be a real number. We can prove by using the continuity of $\ln n$ function that $\;\lim\limits_{n\to\infty}\dfrac{\log_an}n=0\;$

However, this problem appears in my problems book quite early right after the definition of $\epsilon$-language definition of limit of a sequence, the reader is supposed not to know anything about continuity.

My question is: Is there any proof for this result in $\epsilon-delta$ language that is more elementary?

Please help me.

Thanks.

To prove that $\displaystyle \begin{align*} \lim_{n \to \infty}{ \frac{\log_a{\left( n \right) }}{n} } = 0 \end{align*}$ we would need to prove that:

For any $\displaystyle \begin{align*} \epsilon > 0 \end{align*}$ there exists an $\displaystyle \begin{align*} N > 0 \end{align*}$ such that $\displaystyle \begin{align*} n > N \implies \left| \frac{\log_a{\left( n \right) }}{n} - 0 \right| < \epsilon \end{align*}$.

So to do this:

$\displaystyle \begin{align*} \left| \frac{\log_a{\left( n \right) }}{n} - 0 \right| &< \epsilon \\
\left| \log_a{\left( n \right) } \right| &< \epsilon \left| n \right| \\
\left| \frac{\ln{ \left( n \right) }}{\ln{ \left( a \right) }} \right| &< \epsilon \left| n \right| \\
\left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \end{align*}$

Now, obviously we are considering what happens for very large $\displaystyle \begin{align*} n \end{align*}$, so it is obvious that there is no reason why we need to consider all $\displaystyle \begin{align*} n > 0 \end{align*}$, so why don't we consider say $\displaystyle \begin{align*} n > \mathrm{e} \implies \ln{ \left( n \right) } > 1 \end{align*}$.

Therefore, if $\displaystyle \begin{align*} n > \mathrm{e} \end{align*}$

$\displaystyle \begin{align*} 1 < \left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
1 &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
\frac{1}{\epsilon \left| \ln{ \left( a \right) } \right| } &< \left| n \right| \\
\left| n \right| &> \frac{1}{\epsilon \left| \ln{ \left( a \right) } \right| } \end{align*}$

So provided that we ensure we start with an $\displaystyle \begin{align*} n > \mathrm{e} \end{align*}$, we can set $\displaystyle \begin{align*} N = \frac{1}{\epsilon \left| \ln{ \left( a \right) } \right| } \end{align*}$ and then you can write the proof.

 

[Evgeny.Makarov]

$\displaystyle \begin{align*} 1 < \left| \ln{ \left( n \right) } \right| &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right| \\
1 &< \epsilon \left| \ln{ \left( a \right) } \right| \left| n \right|\end{align*}$

And what is the relationship between these two lines?

 

[Prove It]

And what is the relationship between these two lines?

What do you mean?

 

[Evgeny.Makarov]

Well, a proof is not a collection of random statements, is it? There must be a logical relationship between statements, for example, all of them should be equivalent or each of them should imply the following one. I am asking about the logical relationship between the two statements in post 5 and why this relationship is sufficient for proving that $\left|\dfrac{\log_a(n)}{n}\right|<\epsilon$ for all $n>\dfrac{1}{\epsilon|\ln(a)|}$.

 

[HOI]

The "logical relationship" between those two lines is
"If a< b< c then a< c".

 

[Evgeny.Makarov]

The "logical relationship" between those two lines is
"If a< b< c then a< c".

OK, but the proof claims that if $n>\dfrac{1}{\epsilon|\ln(a)|}$, then $\left|\dfrac{\log_a(n)}{n}\right|<\epsilon$, and this requires the converse implication. More precisely, $n>\dfrac{1}{\epsilon|\ln(a)|}$ guarantees that $1<\epsilon|\ln(a)|n$, while one needs $\ln(n)<\epsilon|\ln(a)|n$.

 

[Prove It]

OK, but the proof claims that if $n>\dfrac{1}{\epsilon|\ln(a)|}$, then $\left|\dfrac{\log_a(n)}{n}\right|<\epsilon$, and this requires the converse implication. More precisely, $n>\dfrac{1}{\epsilon|\ln(a)|}$ guarantees that $1<\epsilon|\ln(a)|n$, while one needs $\ln(n)<\epsilon|\ln(a)|n$.

Every step is reversible.

 

[Evgeny.Makarov]

Let $a=e$, so $\ln(a)=1$. Are you claiming that if $n>\dfrac{1}{\epsilon}$, then $\dfrac{\ln(n)}{n}<\epsilon$? The assumption only guarantees that $\dfrac{1}{n}<\epsilon$, without $\ln(n)$ in the numerator. For an example, let $n=8>e^2$, then $\ln(n)>2$. Also let $\epsilon=\dfrac{1}{7}$. Then $n>\dfrac{1}{\epsilon}$, but $\dfrac{\ln(n)}{n}>\dfrac{2}{8}=\dfrac{1}{4}>\dfrac{1}{7}=\epsilon$.

 

[Prove It]

I see what your issue is now. Generally with $\displaystyle \begin{align*} \epsilon\delta \end{align*}$ proofs it's assumed that as long as you can find a relationship from a boundary, that every step should be able to be reversed. I'll see if I can prove it the other way as well.

Note, I'm leaving off the absolute values on the terms that are obviously nonnegative.

It seems to work fine for bases $\displaystyle \begin{align*} a > \mathbf{e} \end{align*}$, because then

$\displaystyle \begin{align*} a &> \mathbf{e} \\
\ln{ \left( a \right) } &> \ln{ \left( \mathbf{e} \right) } \\
\ln{ \left( a \right) } &> 1 \\
n \ln{ \left( a \right) } &> n \end{align*}$

and it's well known that $\displaystyle \begin{align*} n > \ln{ \left( n \right) } \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{R} \end{align*}$, so that gives

$\displaystyle \begin{align*} n \ln{ \left( a \right) } &> n > \ln{ \left( n \right) } \\ n\ln {\left( a \right) } &> \ln{ \left( n \right) } \end{align*}$

It also seems to work for bases $\displaystyle \begin{align*} b = \frac{1}{a} \end{align*}$, where $\displaystyle \begin{align*} a \end{align*}$ is as before defined to be $\displaystyle \begin{align*} a > \mathbf{e} \end{align*}$

$\displaystyle \begin{align*} \left| \ln{ \left( b \right) } \right| &= \left| \ln{ \left( \frac{1}{a} \right) } \right| \\
&= \left| - \ln{ \left( a \right) } \right| \\ &= \ln{ \left( a \right) } \end{align*}$

so the same inequality would apply.

I will have to keep thinking about bases in $\displaystyle \begin{align*} \left( \frac{1}{\mathbf{e}} , 1 \right) \cup \left( 1 , \mathbf{e} \right) \end{align*}$.