Proof of inequalities using binomial theorem 
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. 

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