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.

  • $\begingroup$ The fundamentals of mathematics are built on induction (and recursion). If you want to avoid it, you don't have a lot left. That being said, why would you like to avoid it? Induction is beautiful. $\endgroup$ – Arthur Dec 9 '16 at 15:17
  • $\begingroup$ @Arthur, Minimum choices $\implies$ Harder Question. $\endgroup$ – Vidyanshu Mishra Dec 9 '16 at 15:29
  • 1
    $\begingroup$ it's false when n=0 $\endgroup$ – Cato Dec 9 '16 at 15:31

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

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


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

  • 2
    $\begingroup$ Induction is hidden in the binomial theorem... $\endgroup$ – lhf Dec 9 '16 at 16:35
  • 5
    $\begingroup$ @Ihf You can't be that pedantic. Every statement about $\mathbb{N}$ implicitly uses induction. $\endgroup$ – MathematicsStudent1122 Dec 14 '16 at 11:51
  • 1
    $\begingroup$ @lhf not necessairly, you can prove the binomial theorem in many ways not involving induction. A combinatorial proof comes to mind. $\endgroup$ – AspiringMat Dec 20 '16 at 17:01
  • $\begingroup$ @lhf I don't care if induction is here, this is beautiful. $\endgroup$ – Simply Beautiful Art Jan 7 '17 at 13:25

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


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

  • $\begingroup$ This is the approach I would have used. (+1) and Happy Holidays Barry! -Mark $\endgroup$ – Mark Viola Dec 20 '16 at 16:44
  • $\begingroup$ @Dr.MV, thank you, and HH to you too! $\endgroup$ – Barry Cipra Dec 20 '16 at 20:39
  • $\begingroup$ Haha, good old Stirling approximations... $\endgroup$ – Simply Beautiful Art Jan 7 '17 at 13:26

Yet an other point of view ...

Consider the sequence defined by :


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


We have :


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


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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.