I’d like to calculate $$ \lim_{n\to\infty}\frac{\ln n}{n}$$ by using the squeeze theorem.
So I’m looking for a sequence with members that are bigger than $\frac{\ln n}{n}$, but which also converges to zero.
How is it possible to evaluate this sequence without the exponent?
The two properties of $\log$ that we will use are:
From these, you get:
We will show that $n^{1/n}$ is decreasing for $n\geq 3$ and (obviously) bounded below by $1$.
From this (and our assumptions about $\log$) we see that $a_n=\log(n^{1/n})=\frac{\log n}{n}$ is decreasing and bounded below, and hence must converge.
But then the sequence $a_{n^2}$ must converge to the same value, and $a_{n^2}=\frac{2}{n}a_n$. So $a_{n^2}\to 0$, and hence $a_n\to 0$.
Now, we just need to prove that $(n+1)^{1/(n+1)}\leq n^{1/n}$ when $n\geq 3$.
This is equivalent to:
$$n^{n+1}\geq (n+1)^n$$
or:
$$n\geq \left(1+\frac{1}{n}\right)^n=\sum_{k=0}^{n}\binom{n}{k}n^{-k}$$
Now, each of the terms in the sum is know to be $\leq 1$. But you also have that, when $n\geq 3$, that the last two terms $(k=n-1,n)$ add to:
$$\frac{n^2+1}{n^n}$$
which is less than $1$ for $n\geq 3$. So we have:
$$\left(1+\frac{1}{n}\right)^n\leq 1+\sum_{k=0}^{n-2}\binom{n}{k}n^{-k}\leq n,$$
which proves what we needed.
Aside: There is a stronger inequality which comes from $\binom{n}{k}\leq \frac{n^k}{2^{k-1}}$ for $k\geq 1$, giving that $\left(1+\frac1n\right)^n\leq 3.$
An alternative approach is to show that there is a constant $C$ such that for all $n$: $C\leq \sqrt{n}-\log(n)$. From this, we deduce that: $$\frac{\log n}{\sqrt{n}}\leq 1-\frac{C}{\sqrt{n}}$$ and therefore $$0\leq \frac{\log n}{n}\leq \frac{1}{\sqrt{n}}-\frac{C}{n}$$ and thus, by the squeeze theorem, $\frac{\log n}{n}\to 0$.
To prove this, you need the inequality $\log(1+x)\leq x$. Then you can prove by induction that for $n\geq 5$ that $\sqrt{n}-\log(n)<\sqrt{n+1}-\log(n+1)$. Thus, $C$ is the minimum of the values for $n=1,2,3,4,5$.
With $\varphi(x)=\dfrac{\ln x}{x^{1/2}}$ for $x>0$, $\varphi'(x)<0$ for all $x>e^{2}$, so $\varphi(x)\leq\varphi(e^{2})=\text{constant}$, so $\dfrac{\ln x}{x}\leq\varphi(e^{2})\dfrac{1}{x^{1/2}}$, now take $x\rightarrow\infty$.
$y_n: =n^{1/n}$,
Find $\lim_{n \rightarrow \infty } log y_n$.
Step 1:
$(1+x)^n >(n^2/4)x^2,$ $x\ge 0$, $n\ge 2,$
where $x$ real, and $n$ a positive integer.
Proof:
Binomial expansion :
$(1+x)^n > 1 + nx + (n(n-1)/2)x^2+...>$
$(n(n-1)/2)x^2 \ge (n^2/4)x^2 ,$
for $n\ge 2.$
(Since $2(n-1) \ge n$).
Step 2:
$n^{1/n} \le 1+ 2/√n $, $ n \ge 2$
Proof:
Set $x= 2/√n.$
$(1+2/√n)^n \gt (n^2/4 )(2/√n)^2;$
$(1+2/√n) \gt n^{1/n} \gt 1.$
(Note: $n > 1$, then $n^{1/n} \gt 1$)
Step 3:
$1\le \lim_{n \rightarrow \infty} n^{1/n} \le$
$\lim_{n \rightarrow \infty} (1 + 1/√n) =1$.
Step 4:
$y_n =n^{1/n}.$
$\lim_{n \rightarrow \infty} \log(y_n) =0,$
since log is continuos.
