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

Question

I have got an inequality problem which is as follow:

Show that $e^n>\frac{(n+1)^n}{n!}$

I can do it by induction but I have been told to prove it without induction.

My Work:

$$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$ $$e^n>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$ $$e^n>\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$

From here I can't go further.

I shall be thankful if you guys can provide me a complete solution/proof of this inequality. A hint will also work.

Thanks in advance.

Answer 1

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

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

Answer 2

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

Answer 3

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 $(n+1)^n/n!\lt e^n$

Answer 4

Yet an other point of view ...

Consider the sequence defined by :

$$u_n=\frac{e^nn!}{(n+1)^n}$$

and compute the ratio of two consecutive terms (here $n\ge 1$) :

$$\frac{u_{n}}{u_{n-1}}=e\frac{n!}{(n-1)!}\frac{n^{n-1}}{(n+1)^n}=e\left(\frac{n}{n+1}\right)^n$$

We have :

$$\ln\left(\frac{u_n}{u_{n-1}}\right)=1+n\ln\left(\frac{n}{n+1}\right)=1-n\ln\left(1+\frac{1}{n}\right)$$

We know that $\forall x>-1,\ln(1+x)\le x$. Hence :

$$\ln\left(\frac{u_n}{u_{n-1}}\right)=n\left(\frac{1}{n}-\ln\left(1+\frac{1}{n}\right)\right)>0$$

This show that the sequence $(u_n)_{n\ge0}$ is (stricly) increasing. Since $u_0=1$, the proof is complete.

Answer 5

Beware: overkill. The given inequality is equivalent to $n+\log\Gamma(n+1)>n \log(n+1)$, so it is enough to show that $1+\psi(x+1)\geq \frac{x}{x+1}+\log(1+x)$ holds for any $x\geq 0$, which is equivalent to $$ \forall x\geq 0,\qquad \psi(x+2) \geq \log(x+1).\tag{A} $$ On the other hand, if we start with the Weierstrass product for the $\Gamma$ function $$ \Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{z/n}$$ we instantly get: $$ \psi(z+1) = -\gamma+\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{z+n}\right) = \sum_{n\geq 1}\left[\log(n+1)-\log(n)-\frac{1}{n+z}\right]\tag{B} $$ and by the concavity of the logarithm function we have $-w>\log(1-w)$ in a right neighbourhood of the origin, from which: $$\begin{eqnarray*} \psi(x+2)&\geq&\sum_{n\geq 1}\left[\log(n+1)-\log(n)+\log(n+x)-\log(n+x+1)\right]\\&=&\sum_{n\geq 1}\left[\log\left(1+\frac{x}{n}\right)-\log\left(1+\frac{x}{n+1}\right)\right]\\&=&\log(x+1)\tag{C} \end{eqnarray*}$$ by telescopic series.