Prove that the inequality $$\left(1+\frac{1}{n}\right)^n > \frac{9}{4}$$ holds for all integers $n$ beyond a certain point.
I must show that it is true for all $n>3$. but I am having difficult time showing $P(n+1)$ is true using induction. The book give hint: $$[1+(n+1)^{-1}]^{n+1} = \left[\frac{n+2}{n+1}\right]\left[\frac{n+1}{n}\right]^n \left[1-(n+1)^{-2}]^n\right]$$
We can write $$\frac{(n+1)^n}{n^n}$$ By binomial theorem, $$=\frac{{n\choose 0}n^n+{n\choose 1}n^{n-1}+\cdots+{n\choose n}n^0}{n^n}$$ $$={n\choose 0}+{n\choose 1}n^{-1}+\cdots+{n\choose n}n^{-n}$$ $$=1+\frac{n!}{n(n-1)!(1)!}+\frac{n!}{n^2(n-2)!(2)!}+\cdots+\frac{n!}{n^n(0)!(n)!}$$ $$=1+1+\frac{n-1}{2n}+\cdots+\frac{1}{n^n}$$ As you can see, $$1+1+\frac{n-1}{2n}>\frac{9}{4}\forall n>3,n\in\mathbb N$$ (as $\frac{n-1}{2n}$ is strictly increasing for all positive numbers and $2+\frac{3-1}{2(3)}>\frac{9}{4}$) $$\implies 2+\frac{n-1}{2n}>\frac{9}{4}$$ The first three terms are already greater than $\frac{9}{4}$ and rest of the terms are positive, hence proved.
Alternatively, by induction
Base Step
$P(4)$ is true for $n=4$ as$$\frac{19}{8}>\frac{9}{4}$$
Inductive Step
Let $P(k)$ be true.
Then $$P(k+1)\implies 2+\frac{k}{2k+2}>\frac{9}{4}$$ We have to prove $P(k+1)$ is true $$\frac{k}{2k+2}>\frac{1}{4}$$ $$4k>2k+2$$ $$k>1$$ This is obviously true as we have taken $k>3$.
We now have the first three terms always greater than $\frac{9}{4}$. Since rest all terms are positive, hence proved by induction.