Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$ I need help understanding this:
Prove that $$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$$

$$(\sqrt n)^\frac{1}{n}\geq 
(\sqrt1)^\frac{1}{n}=1, \forall n\in\Bbb N$$
  By binomial theorem we get for $n\geq 2$
$$n=((\sqrt n)^\frac{1}{n})^n=[1+((\sqrt n)^\frac{1}{n}-1)]^n=\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k$$
  $$\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2=1+\frac{n(n-1)}{2}(\sqrt n^\frac{1}{n}-1)^2$$
  $$\Rightarrow (\sqrt n^\frac{1}{n}-1)^2\leq \frac{2}{n}$$
  $$\Rightarrow \sqrt n^\frac{1}{n}\leq 1+\frac{\sqrt 2}{\sqrt n}$$
  $1+\frac{\sqrt 2}{\sqrt n}$ approaches $1$ for $n \rightarrow \infty$ so by the sandwich theorem we get $\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$

I don't understand this part: 
$$
\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2.$$ Could someone explain how they got that on the RHS?
 A: Since $n^{1/n}\ge 1$, then all of the terms in the binomial expansion
$$\left(1+(\sqrt[n]{n}-1)\right)^n=\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k$$
are non-negative.  Therefore, we have for any $m\le n$
$$\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k\ge \sum_{k=0}^m\binom{n}{k}(\sqrt[n]{n}-1)^k$$
Taking $m=2$ and $n\ge 2$ yields
$$\begin{align}
\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k&\ge \sum_{k=0}^2\binom{n}{k}(\sqrt[n]{n}-1)^k\\\\
&=1+n(\sqrt[n]{n}-1)+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2\\\\
&\ge 1+\frac{n(n-1)}{2}(\sqrt[n]{n}-1)^2
\end{align}$$
where the last inequality is true since $n(\sqrt[n]{n}-1)\ge 0$.
A: They are truncating the binomial expansion, discarding positive terms: if
$$
r=r_0+r_1+r_2+\dots+r_n
$$
and $r_i\ge0$ for $i=0,1,\dots,n$, then $r\ge r_0+r_2$.

It can be done more easily. Set $(\sqrt{n})^{1/n}=1+a_n$, so
$$
\sqrt{n}=(1+a_n)^{n}\ge1+na_n
$$
by Bernoulli’s inequality. Hence
$$
a_n\le\frac{\sqrt{n}-1}{n}=\frac{1}{\sqrt{n}}-\frac{1}{n}
$$
so
$$
(\sqrt{n})^{1/n}=1+a_n\le 1+\frac{1}{\sqrt{n}}-\frac{1}{n}
$$
and the squeeze theorem allows to conclude.
A: Set $l =\lim_{n\to\infty}n^\frac{1}{n}$
$$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=\sqrt{\lim_{n\to\infty}n^\frac{1}{n}}=\sqrt{l}$$
For the limit of the sequence $n^\frac{1}{n}$ get the sub-sequence $(2n)^\frac{1}{2n}$
So: 
$l^2=(\lim (2n)^\frac{1}{2n})^2=\lim (2n)^\frac{1}{n}=\lim 2^\frac{1}{n}. n^\frac{1}{n}=\lim 2^\frac{1}{n}.\lim n^\frac{1}{n}=l$ 
$l^2=l$ and as $n^\frac{1}{n}\gt1$ we have $l=1$
And $\sqrt{\lim_{n\to\infty}n^\frac{1}{n}}=\sqrt{1}=1$
A: That proof seems a little complicated; I'll do it a different way.  Let's write $(\sqrt n)^{1/n} = 1+a_n.$ We know each $a_n>0.$ For $n\ge 2,$ the binomial theorem shows
$$\sqrt n= (1+a_n)^{n} = 1^n\cdot a_n^0 + n\cdot1^{n-1}\cdot a_n^1 + [n(n-1)/2]\cdot 1^{n-2}\cdot a_n^2 + \cdots > [n(n-1)/2]a_n^2.$$
Therefore 
$$\frac{\sqrt n}{n(n-1)/2} > a_n^2.$$
Since the expression on the left $\to 0,a_n^2 \to 0,$ which implies $a_n \to 0.$ It follows that $(\sqrt n)^{1/n} \to 1.$
A: A Different Approach
Because $\left(1+\frac1n\right)^{n+1}$ is decreasing, for $n\ge3$,
$$
\begin{align}
\frac{(n+1)^n}{n^{n+1}}
&=\frac1{n+1}\left(1+\frac1n\right)^{n+1}\\
&\le\frac14\cdot\frac{256}{81}\\[6pt]
&\lt1
\end{align}
$$
Thus, for $n\ge3$,
$$
(n+1)^{\frac1{n+1}}\lt n^{\frac1n}
$$
That is, $n^{\frac1n}$ is decreasing and bounded below by $1$. So $\lim\limits_{n\to\infty}n^{\frac1n}$ exists.
$$
\begin{align}
\lim_{n\to\infty}n^{\frac1n}
&=\lim_{n\to\infty}(2n)^{\frac1{2n}}\\
&=\lim_{n\to\infty}2^{\frac1{2n}}\lim_{n\to\infty}n^{\frac1{2n}}\\
\end{align}
$$
Therefore,
$$
\begin{align}
\lim_{n\to\infty}n^{\frac1{2n}}
&=\lim_{n\to\infty}2^{\frac1{2n}}\\[6pt]
&=1
\end{align}
$$

Bernoulli's Inequality
Bernoulli's Inequality says
$$
\begin{align}
\lim_{n\to\infty}\sqrt{n}^{1/n}
&=\lim_{n\to\infty}\left(1+\left(\sqrt{n}-1\right)\right)^{1/n}\\[3pt]
&\le\lim_{n\to\infty}\left(1+\tfrac1n\left(\sqrt{n}-1\right)\right)\\[6pt]
&=1
\end{align}
$$
Since $\sqrt{n}^{1/n}\ge1$, the Squeeze Theorem says
$$
\lim_{n\to\infty}\sqrt{n}^{1/n}=1
$$
A: $$\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k= 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2+\left[{{n}\choose {1}}(\sqrt n^\frac{1}{n}-1)^1+\sum_{k=3}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\right]$$
and see that:
$${{n}\choose {1}}(\sqrt n^\frac{1}{n}-1)^1+\sum_{k=3}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k \ge 0$$
because 
$$\sqrt{n}^\frac{1}{n}-1 \ge0$$
A: I'd like to offer another method, using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$
$$n-1=(n^{\frac{1}{2n}})^{2n}-1=\\(n^{\frac{1}{2n}}-1)(n^{\frac{1}{2n}(2n-1)}+n^{\frac{1}{2n}(2n-2)}+...+n^{\frac{1}{2n}2}+n^{\frac{1}{2n}}+1)\geq$$
using AM-GM
$$\geq(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1)}\cdot n^{\frac{1}{2n}(2n-2)}\cdot ...\cdot n^{\frac{1}{2n}2}\cdot n^{\frac{1}{2n}}}+1\right)=\\
(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1+2n-2+...+1)}}+1\right)=
\\(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}\frac{(2n-1)2n}{2}}}+1\right)=\\
(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt{n}+1\right)$$
Also $\ln{x}\leq x-1,x>0$ which means $$\ln{n^{\frac{1}{2n}}}=\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1$$
Altogether 
$$\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1\leq \frac{n-1}{(2n-1)\sqrt{n}+1}<\frac{1}{2\sqrt{n}}$$
A: Let $\lim_{n\to\infty} n^{1/2n} = L$
Taking logarithms, we get ;
$log(L) =$$\lim_{n\to\infty}$$\frac{1}{2n}$$log(n)$
Applying $LH$ Rule :
$log(L) =$$\lim_{n\to\infty}$$\frac{1}{2n}=0$
Hence, $L=e^0=1$
