Proof that $\lim_{n\to\infty} (n+1)^{1/n} = 1$ How would I go about showing that
$$
\lim_{n\to\infty} (n+1)^{1/n} = 1
$$
without using L'Hopital's rule? Through writing a MATLAB code, I confirmed that it is $1$ - I just need to formally prove it.
 A: You can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then
$$
\lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}
$$
(see this question).
We have that
$$
\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{n+2}{n+1}=1
$$
and
$$
\lim_{n\to\infty}a_n^{1/n}=1.
$$
A: $$(n+1)^{1/n}=e^{(1/n) \cdot \ln{n+1}}$$
Now you need to show that the thing in the exp converges to 0
Lemma : $ln(x)<\sqrt{x}$
Proof : the function $f(x)=\sqrt{x} - ln(x)$ is increasing (you can see it instantly after deriving) from $x=\sqrt2$. In $x=\sqrt2$ the inequality from the lemma is already true so it will be true for all $x$ greater than $\sqrt2$
In our case, for $n>2$: $$0<\ln{n+1}<\sqrt{n+1}$$
Thus : $$0<\frac{\ln{n+1}}{n} < \frac{\sqrt{n+1}}{n}$$
This means the limit of the part in the exp is 0 by comparison. As exp is a continuous function, it means :
$$\lim_{n\to \infty}{(n+1)}^{1/n} = e^0 = 1$$
A: 1.Lemma 1.
$$\lim_{n\to\infty}\frac{a^n}{n}=\infty, (a>1)$$
proof:
$$a=1+\gamma, \hbox{where}\ \gamma>0$$
Using Newton's binomial formula:
$$a^n=(1+\gamma)^n>1+n\gamma+\frac{n(n-1)\gamma^2}{2}\ ,(n>2)$$ 
We arrive at:
$$\lim_{n\to\infty}\frac{a^n}{n}=\lim_{n\to\infty}\frac{1+n\gamma+\frac{n(n-1)\gamma^2}{2}}{n}=\infty$$
2.Lemma 2.
From lemma 1, we can observe that:
$$\lim_{n\to\infty}\frac{n}{a^n}=0\ ,(a>1)$$
3.Lemma 3.
In lemma 2, let 
$$t=a^n>1$$
$$n=\log_a t$$
for any t>1, and a>1, there exist an(unique) $$n\in\mathbb{N}$$, such that
$$a^n\le t<a^{(n+1)}$$
$$n\le \log_a t<(n+1)$$
Taking the two inequality together, we have
$$\frac{n}{a^{(n+1)}} < \frac{\log_a t}{t}< \frac{(n+1)}{a^n}$$
Taking limits on both sides and using the sequeeze theorem, we have
$$\lim_{t\to\infty}\frac{\log_a t}{ t}=0$$
4.Lemma 4.
Taking the discrete sequence of t in lemma 3, we have:
$$\lim_{n\to\infty}\frac{\log_a n}{n}=0$$
5.Theorem
Since the function $$f(x)=a^x$$ is a continuous function, so $$
\lim_{x\to 0}a^x=a^0=1$$
Appling this to lemma 4, we get
$$\lim_{n\to\infty}a^{\frac{\log_a n}{n}}=\lim_{n\to\infty}n^{\frac{1}{n}}=1$$
At last
$$\lim_{n\to\infty}(n+1)^{\frac{1}{n}}=\lim_{n\to\infty}\left(\frac{n+1}{n}\times n\right)^{\frac{1}{n}}=1\times 1=1$$
