Proof of inequalities using binomial theorem

Question

How do they achieve $$1+1+\frac{1}{2}\left(1-\frac{1}{n}\right)?$$ If I consider the first three terms of the binomial expansion, I get the first two terms, but can't factor the third to match this.

Answer 1

\begin{align} & \frac{n(n-1)}{2}\left(\frac1n\right)^2 \\ =& \frac{n(n-1)}{2} \; \frac1{n^2} \\ =& \frac12\;\frac{n-1}{n} \\ =& \frac{1}{2}\left(1-\frac{1}{n}\right) \end{align}

Answer 2

$(1+x)^n = 1 + n x + \frac {n(n-1)}{2} x^2 + \cdots$

All the terms in the "$\cdots$" are positive if $x$ is positive.

If $n \ge 2$

$(1+x)^n \ge 1 + n x + \frac {n(n-1)}{2} x^2 $

$(1+\frac {1}{n})^n$ replace $x$ above with $\frac 1n$

$(1+\frac 1n)^n \ge 1 + n \frac 1n + \frac {n(n-1)}{2} (\frac 1n)^2 $

and distribute

$(1+\frac 1n)^n \ge 1 + 1 + \frac {n^2-n}{2n^2} = 2+\frac {n-1}{2n} = 2+\frac 12 - \frac {1}{2n} $