Answer
(a) We obtain the following:
$\mathop \smallint \limits_1^n \ln x{\rm{d}}x \lt \ln n! \lt \mathop \smallint \limits_1^{n + 1} \ln x{\rm{d}}x$
(b) We show that
$\dfrac{{{n^n}}}{{{{\rm{e}}^{n - 1}}}} \lt n! \lt \dfrac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{{{\rm{e}}^n}}}$, ${\ \ \ }$ for $n \gt 1$
(c) We prove that
$\mathop {\lim }\limits_{n \to + \infty } \dfrac{{\sqrt[n]{{n!}}}}{n} = \dfrac{1}{{\rm{e}}}$
Work Step by Step
(a) 1. Referring to the graph on the left of Figure Ex-32, we see that the width of the subintervals is $\Delta n = 1$. Thus, the area of the rectangles is
${\rm{Area1}} = \ln 1 + \ln 2 + \ln 3 + ... + \ln \left( {n - 1} \right)$
${\rm{Area1}} = \ln \left( {2\cdot3\cdot ... \cdot\left( {n - 1} \right)} \right) = \ln \left( {\left( {n - 1} \right)!} \right)$
The graph depicts the fact that
$\mathop \smallint \limits_1^n \ln x{\rm{d}}x \lt {\rm{Area1}}$
Or
$\mathop \smallint \limits_1^n \ln x{\rm{d}}x \lt \ln \left( {\left( {n - 1} \right)!} \right)$
Since $\ln \left( {\left( {n - 1} \right)!} \right) \lt \ln n!$, consequently
(1) ${\ \ \ \ \ \ \ }$ $\mathop \smallint \limits_1^n \ln x{\rm{d}}x \lt \ln n!$
2. Referring to the graph on the right of Figure Ex-32, we also see that the width of the subintervals is $\Delta n = 1$. Thus, the area of the rectangles is
${\rm{Area2}} = \ln 2 + \ln 3 + \ln 4 + ... + \ln n$
${\rm{Area2}} = \ln \left( {2\cdot3\cdot4\cdot ...\cdot n} \right) = \ln n!$
The graph depicts the fact that
$\mathop \smallint \limits_1^{n + 1} \ln x{\rm{d}}x \gt {\rm{Area2}}$
Or
(2) ${\ \ \ \ \ \ \ }$ $\mathop \smallint \limits_1^{n + 1} \ln x{\rm{d}}x \gt \ln n!$
Combining equation (1) and equation (2), we obtain
(3) ${\ \ \ \ \ \ \ }$ $\mathop \smallint \limits_1^n \ln x{\rm{d}}x \lt \ln n! \lt \mathop \smallint \limits_1^{n + 1} \ln x{\rm{d}}x$
(b) Evaluate the integrals:
From the Table of Integrals, we know that
$\mathop \smallint \limits_{}^{} \ln u{\rm{d}}u = u\ln u - u + C$
So,
$\mathop \smallint \limits_1^n \ln x{\rm{d}}x = n\ln n - n + 1 = 1 - n + \ln {n^n}$
$\mathop \smallint \limits_1^{n + 1} \ln x{\rm{d}}x = \left( {n + 1} \right)\ln \left( {n + 1} \right) - n = - n + \ln {\left( {n + 1} \right)^{n + 1}}$
Substituting these results into equation (3) gives
$1 - n + \ln {n^n} \lt \ln n! \lt - n + \ln {\left( {n + 1} \right)^{n + 1}}$
Taking the exponentials, we get
${{\rm{e}}^{1 - n + \ln {n^n}}} \lt {{\rm{e}}^{\ln n!}} \lt {{\rm{e}}^{ - n + \ln {{\left( {n + 1} \right)}^{n + 1}}}}$
$\dfrac{{{{\rm{e}}^{\ln {n^n}}}}}{{{{\rm{e}}^{n - 1}}}} \lt {{\rm{e}}^{\ln n!}} \lt \dfrac{{{{\rm{e}}^{\ln {{\left( {n + 1} \right)}^{n + 1}}}}}}{{{{\rm{e}}^n}}}$
Since ${{\rm{e}}^{\ln x}} = x$, so
$\dfrac{{{n^n}}}{{{{\rm{e}}^{n - 1}}}} \lt n! \lt \dfrac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{{{\rm{e}}^n}}}$
Hence,
$\dfrac{{{n^n}}}{{{{\rm{e}}^{n - 1}}}} \lt n! \lt \dfrac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{{{\rm{e}}^n}}}$, ${\ \ \ }$ for $n \gt 1$
(c) From part (b), we obtain
$\dfrac{{{n^n}}}{{{{\rm{e}}^{n - 1}}}} \lt n! \lt \dfrac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{{{\rm{e}}^n}}}$
Or
$\dfrac{{{\rm{e}}{n^n}}}{{{{\rm{e}}^n}}} \lt n! \lt \dfrac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{{{\rm{e}}^n}}}$
Take the $n$th root:
${\left( {\dfrac{{{\rm{e}}{n^n}}}{{{{\rm{e}}^n}}}} \right)^{1/n}} \lt \sqrt[n]{{n!}} \lt {\left( {\dfrac{{{{\left( {n + 1} \right)}^{n + 1}}}}{{{{\rm{e}}^n}}}} \right)^{1/n}}$
$\dfrac{{n{{\rm{e}}^{1/n}}}}{{\rm{e}}} \lt \sqrt[n]{{n!}} \lt \dfrac{{{{\left( {n + 1} \right)}^{1 + \dfrac{1}{n}}}}}{{\rm{e}}}$
$\dfrac{{n{{\rm{e}}^{1/n}}}}{{\rm{e}}} \lt \sqrt[n]{{n!}} \lt \dfrac{{\left( {n + 1} \right){{\left( {n + 1} \right)}^{\dfrac{1}{n}}}}}{{\rm{e}}}$
Divide by $n$:
$\dfrac{{{{\rm{e}}^{1/n}}}}{{\rm{e}}} \lt \dfrac{{\sqrt[n]{{n!}}}}{n} \lt \dfrac{{\left( {1 + \dfrac{1}{n}} \right){{\left( {n + 1} \right)}^{\dfrac{1}{n}}}}}{{\rm{e}}}$
Take the limits:
$\mathop {\lim }\limits_{n \to \infty } \dfrac{{{{\rm{e}}^{1/n}}}}{{\rm{e}}} \lt \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{{n!}}}}{n} \lt \mathop {\lim }\limits_{n \to \infty } \dfrac{{\left( {1 + \dfrac{1}{n}} \right){{\left( {n + 1} \right)}^{\dfrac{1}{n}}}}}{{\rm{e}}}$
As $n \to \infty $, we get $\dfrac{1}{n} \to 0$. Therefore, $\mathop {\lim }\limits_{n \to \infty } {{\rm{e}}^{1/n}} = 1$, and $\mathop {\lim }\limits_{n \to \infty } {\left( {n + 1} \right)^{\dfrac{1}{n}}} = 1$.
It follows that
$\dfrac{1}{{\rm{e}}} \le \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sqrt[n]{{n!}}}}{n} \le \dfrac{1}{{\rm{e}}}$
By the Squeezing Theorem for Sequences (Theorem 9.1.5), we obtain
$\mathop {\lim }\limits_{n \to + \infty } \dfrac{{\sqrt[n]{{n!}}}}{n} = \dfrac{1}{{\rm{e}}}$
