# Proving $\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}}$ by induction for all $n> 2$.

I am trying to prove

$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}} \text{ for all }n > 2.$$

Here is the original source (Problem 1B, on page 12 of PDF)

Can this be proved by induction?

The base step $n=3$ is proved: $\frac {27}{e^2} < 6 < \frac{256}{e^3}$ (since $e^2 > 5$ and $e^3 < 27$, respectively).

I can assume the case for $n=k$ is true: $\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}$.

For $n=k+1$, I am having trouble:

\begin{align} (k+1)!&=(k+1)k!\\&>(k+1)\frac{k^k}{e^{k-1}}\\&=e(k+1)\frac{k^k}{e^{k}} \end{align} Now, by graphing on a calculator, I found it true that $ek^k >(k+1)^k$ (which would complete the proof for the left inequality), but is there some way to prove this relation?

And for the other side of the inequality, I am also having some trouble: \begin{align} (k+1)!&=(k+1)k!\\&<(k+1)\frac{(k+1)^{k+1}}{e^{k}}\\&=\frac{(k+1)^{k+2}}{e^k}\\&<\frac{(k+2)^{k+2}}{e^k}. \end{align} I can't seem to obtain the $e^{k+1}$ in the denominator, needed to complete the induction proof.

• just curious, what level this exam is -to enter PhD program or some qual? – Alex Feb 15 '15 at 13:47
• @Alex According to the link provided in my question, it's UC Berkeley's "preliminary exam", which apparently their students must pass before they can continue their Ph.D program and eventually take their oral qualifying exam. – Cookie Feb 15 '15 at 17:52
• Also, there is a solution here math.berkeley.edu/sites/default/files/pages/… (page 12 of PDF), but the solution uses integration. I understand that method works, but I wanted to also prove this by induction. Induction must be possible to use here... – Cookie Feb 15 '15 at 17:55
• so it's a prelim to qual exam? – Alex Feb 16 '15 at 0:05
• It's a prelim required to stay in the Ph.D program. The student has two chances (or maybe three based on appeal) to pass the exam, or else the student is dismissed. – Cookie Feb 16 '15 at 0:27

Let's try an inductive proof from the original inquality for $n$, let's prove for $n+1$

$$\frac{n^n}{e^{n-1}}<n!<\frac{(n+1)^{n+1}}{e^{n}}$$

Okay, multiply both sides by $n+1$. At least the middle is correct

$$(n+1)\frac{n^n}{e^{n-1}}<(n+1)!<\frac{(n+1)^{n+2}}{e^{n}}$$

and we have to make the left and right sides look more appropriate

$$\left(\color{red}{\frac{n}{n+1}} \right)^\color{red}{n}\frac{(n+1)^{n+1}}{e^{n-1}}<(n+1)!<\frac{(n+2)^{n+2}}{e^{n}} \left(\color{blue}{\frac{n+1}{n+2}}\right)^{\color{blue}{n+2}}$$

Our induction is complete if we can prove two more inequalities:

$$\frac{1}{e} < \left(\frac{n}{n+1} \right)^n \text{ and } \left(\frac{n+1}{n+2}\right)^{n+2}< \frac{1}{e}$$

These two inequalities can be combined into one and we can take reciprocals. At least it is well-known.

$$\bigg(1 + \frac{1}{m}\bigg)^{m+1}> \mathbf{e} > \bigg(1 + \frac{1}{n} \bigg)^n$$

This is true for any $m, n \in \mathbb{N}$.

You shave off a little bit too much when you said $(k+1)^{k+2} < (k+2)^{k+2}$. Instead, you needed the more delicate:

$$(k+1)^{k+2} < \frac{1}{e} (k+2)^{k+2}$$

If you know1 that: $$a_n=\left(1+\frac{1}{n}\right)^n,\qquad b_n = \left(1+\frac{1}{n}\right)^{n+1}$$ give to sequences converging towards $e$, where $\{a_n\}$ is increasing while $\{b_n\}$ is decreasing, consider that:

$$n = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right),\tag{1}$$ so: $$n! = \prod_{m=2}^{n} m = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right) = \prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}\tag{2}$$ and your inequality trivially follows.

1) If you are not aware of such a classical result, then prove it by induction. It is rather easy.

• I think that it is not as easy as you think. – marty cohen Feb 15 '15 at 4:09
• @martycohen: it is also possible to use AM-GM in order to prove that $\{a_n\}$ is increasing and $\{b_n\}$ is decreasing. I think there are many questions/answers on MSE devoted to such well-known fact. – Jack D'Aurizio Feb 15 '15 at 4:12
• Here, for example: math.stackexchange.com/questions/389793/… – marty cohen Feb 15 '15 at 4:18

Proof: We will prove the inequality by induction. Since $e^2 > 5$ and $e^3 < 27$, we have $$\frac {27}{e^2} < 6 < \frac{256}{e^3}.$$ Thus, the statement for $n=3$ is true. The base step is complete.

For the induction step, we assume the statement is true for $n=k$. That is, assume $$\frac{k^k}{e^{k-1}}<k!<\frac{(k+1)^{k+1}}{e^{k}}.$$

We want to prove that the statement is true for $n=k+1$. It is straightforward to see for all $k > 2$ that $\left(1+\frac 1k \right)^k < e < \left(1+\frac 1k \right)^{k+1}$; this algebraically implies $$\left(\frac k{k+1} \right)^{k+1} < \frac 1e < \left(\frac k{k+1} \right)^k. \tag{*}$$ A separate induction proof for the left inequality of $(*)$ establishes $\left(\frac{k+1}{k+2} \right)^{k+2}<\frac 1e$. We now have \begin{align} (k+1)! &= (k+1)k! \\ &< (k+1) \frac{(k+1)^{k+1}}{e^k} \\ &= \frac{(k+1)^{k+2}}{e^k} \\ &= \frac{(k+1)^{k+2}}{e^k} \left( \frac{k+2}{k+2} \right)^{k+2} \\ &= \frac{(k+2)^{k+2}}{e^k} \left( \frac{k+1}{k+2} \right)^{k+2} \\ &< \frac{(k+2)^{k+2}}{e^k} \frac 1e \\ &= \frac{(k+2)^{k+2}}{e^{k+1}} \end{align} and \begin{align} (k+1)! &= (k+1)k! \\ &> (k+1) \frac{k^k}{e^{k-1}} \\ &= (k+1) \frac{k^k}{e^{k-1}} \left(\frac{k+1}{k+1} \right)^k \\ &= \frac{(k+1)^{k+1}}{e^{k-1}} \left( \frac k{k+1} \right)^k \\ &> \frac{(k+1)^{k+1}}{e^{k-1}} \frac 1e \\ &= \frac{(k+1)^{k+1}}{e^k}. \end{align} We have established that the statement $$\frac{(k+1)^{k+1}}{e^k}<(k+1)!<\frac{(k+2)^{k+2}}{e^{k+1}}$$ for $n=k+1$ is true. This completes the proof.

Some times it is easier for such an induction if we shift the sequence by one step, so the basis of the expressions is nicer for algebraic manipulations.
Let's rewrite your sequence of inequalities as
$$\displaystyle {n! \over e}<{(n+1)^{n+1} \over e^{n+1}} < {(n+1)! \over e} <{(n+2)^{n+2} \over e^{n+2}}< {(n+2)! \over e} \tag 1$$
and for simpler references below as $$a_0 \quad < \quad b_0 \quad <\quad a_1 \quad <\quad b_1 \quad < \quad a_2 \tag 2$$

Then we ask: does from $a_0<b_0<a_1$ follow that $a_1<b_1<a_2$ ?

Of course $a_1 = (n+1) \cdot a_0$ and so it might be useful to define $b_0$ as a fraction in the interval of $a_0$ and $a_1$: $$b_0 = (n+1)q_1 \cdot a_0 \text{ where } q_1<1 \tag {3.1 }$$. Consequently, define $$b_1 = (n+2)q_2 \cdot a_1 \text{ where also } q_2<1 \tag {3.2 }$$ Here the inequality $q_2 < 1$ is not known but expected and if this can be shown by induction from $q_1$ this would solve the problem .

So we start with $$q_1 = {b_0 \over a_0 (n+1)} = {(n+1)^{n} \over e^n n! } \tag {4.1 }$$ and by the beginning of the induction we know, that this is indeed smaller than 1 so $$q_1 < 1 \tag {4.2 }$$

Now we have simply $$q_2 = {b_1 \over a_1 (n+2)} = {(n+2)^{n+1} \over e^{n+1} (n+1)! } \tag {4.3 }$$ Next we consider the systematic progression in the sequence of $q_1,q_2,q_3,...$. To begin we determine the ratio $r_2={q_2 \over q_1}$ . We find $$\begin{eqnarray} r_2&=&{q_2 \over q_1}& =&{ {(n+2)^{n+1} \over e^{n+1} (n+1)! } \over {(n+1)^{n} \over e^{n} (n)! } } \\ &&&=& {(n+2)^{n+1} e^{n} (n)! \over e^{n+1} (n+1)! (n+1)^{n}} \\ &&&=& {(n+2)^{n+1} \over e (n+1)^{n+1}} \\ &&&=& \left({n+2 \over n+1}\right)^{n+1} \cdot \frac 1e \\ r_2&=&{q_2 \over q_1}& =& \left( 1 + {1 \over n+1} \right)^{n+1} \cdot \frac 1e \end{eqnarray} \tag {5.1 }$$

It is now needed to recognize/remember from the definition of $e$, that the last expression is smaller than 1 and that for $n \to \infty$ approximates monotonically 1.
Also we see by the expansion of the binomial-series $(1+1/x)^x=1+1+1/2+...$ in the general case that for $x>2$ this is greater than $2$ so $${2 \over e} \approx 0.73 < r_2 < 1 \tag{5.2}$$

From this we know now, that $q_2$ is not only smaller than $1$ but also smaller than $q_1$ but$q_{n+1} \to q_n$ for increasing $n$.

So we have $$\begin{eqnarray} &q_2 &=& q_1 \cdot r_2 < q_1 < 1 \\ \to& b_1 &<& (n+2) a_1 = a_2 \\ \to &a_1 &<& b_1 <a_2 \end{eqnarray} \tag {6 }$$ which we wanted to show.

Since several answers have already proven the result via induction, I see no harm in recording an additional proof that does not use induction.

We claim for all $n\ge 1$, $$\int_{0}^{1} (x\log(x))^n \, dx = \frac{(-1)^n n!}{(n+1)^{n+1}}$$ This can be shown using differentiation under the integral sign; set $k=n$ below: $$f(n) = \int_{0}^{1} x^n \, dx = \frac{1}{n+1} \implies \frac{d^k}{dn^k} f(n) = \int_{0}^{1} x^n (\log(x))^k \, dx = \frac{(-1)^{k} k!}{(n+1)^{k+1}}$$

Since $x \log(x)$ does not change sign, we have $$\int_{0}^{1} |x\log(x)|^n \, dx = \frac{n!}{(n+1)^{n+1}}$$ From this expression, we see that the given inequality is equivalent to $$\frac{1}{(n+1)e^n} < \int_{0}^{1} |x\log(x)|^n \, dx < \frac{1}{e^n}$$

A simple computation shows $|x\log(x)|^n$ attains a maximum of $e^{-n}$ at $x=e^{-1}$, so the upper bound is evident. To see the lower bound, note $|x\log(x)|^n$ is a concave function, so its graph encloses the triangle with vertices $(0,0)$, $(1,0)$, and $(e^{-n}, e^{-1})$. This triangle has area $e^{-n}/2 > e^{-n}/(n+1)$ since $n>2$.