# Number Theory using an induction proof

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: $$[1+(n+1)^{-1}]^{n+1} = \left[\frac{n+2}{n+1}\right]\left[\frac{n+1}{n}\right]^n \left[1-(n+1)^{-2}]^n\right]$$

• What have you tried so far? Can you write out the steps up to the point you get stuck? – graydad Feb 26 '15 at 15:31
• You can use binomial theorem if you can't get induction to work – AvZ Feb 26 '15 at 15:34

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.
$$P(4)$$ is true for $$n=4$$ as$$\frac{19}{8}>\frac{9}{4}$$
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.