TITLE: Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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.