Show that $e^n>\frac{(n+1)^n}{n!}$ without using induction.

Solution 1:

$$n!e^n\ge\sum_{k=0}^n\frac{n!}{k!}n^k\ge\sum_{k=0}^n\binom nkn^k=(n+1)^n$$

Solution 2:

From $\int\ln x\,dx=x\ln x-x+C$, we get

$$\int_1^{n+1}\ln x\,dx=(n+1)\ln(n+1)-n$$

But since $\ln x$ is strictly increasing, we have

$$\int_1^{n+1}\ln x\,dx\lt\ln2+\ln3+\cdots+\ln n+\ln(n+1)=\ln(n!)+\ln(n+1)$$

It follows that

$$n\ln(n+1)-n\lt\ln(n!)$$

which exponentiates to $(n+1)^n/e^n\lt n!$, or $(n+1)^n/n!\lt e^n$

Solution 3:

$$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$ $$e^n>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$ $$e^n>\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$ $$e^n>n^n\left[\frac{1}{n!}+\frac{1}{n(n-1)!}+\frac{1}{n^2(n-2)!}...+\frac{1}{n^{n-1}}+\frac{1}{n^n}\right] $$ $$e^n>\frac{n^n}{n!}\left[1+\frac{1}{n}n+\frac{1}{n^2}n(n-1)+\frac{1}{n^3}n(n-1)(n-2)...+\frac{n!}{n^n}\right] $$ $\because$ $$n(n-1)>\frac{n(n-1)}{2!}$$and $$n(n-1)(n-2)>\frac{n(n-1)(n-2)}{3!}$$and $$n!>1$$

$\therefore $ $$e^n>\frac{n^n}{n!}\left[1+n\frac{1}{n}+\frac{n(n-1)}{2!}\frac{1}{n^2}+...+\frac{1}{n^n}\right]$$ $$e^n>\frac{n^n}{n!}(1+\frac1n)^n$$ $$e^n>\frac{n^n}{n!}\frac{(n+1)^n}{n^n}$$ $$e^n>\frac{(n+1)^n}{n!}$$