\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
For the left inequality, \(f(n) > 0\) since \(f(n) > \frac {1}{n + 1} > 0\).
For the right inequality, we notice that \begin {align*} f(n) & = \frac {1}{n + 1} + \frac {1}{(n + 1) (n + 2)} + \frac {1}{(n + 1) (n + 2) (n + 3)} + \cdots \\ & < \frac {1}{n + 1} + \frac {1}{(n + 1)^2} + \frac {1}{(n + 1)^3} + \cdots \\ & = \frac {1}{n + 1} \cdot \frac {1}{1 - \frac {1}{n + 1}} \\ & = \frac {1}{(n + 1) - 1} \\ & = \frac {1}{n}. \end {align*}
Hence, \[ 0 < f(n) < \frac {1}{n}. \]
For the left inequality, by grouping consecutive terms, we have \begin {align*} g(n) & = \frac {1}{n + 1} - \frac {1}{(n + 1) (n + 2)} \\ & \phantom {=} + \frac {1}{(n + 1) (n + 2) (n + 3)} - \frac {1}{(n + 1) (n + 2) (n + 3) (n + 4)} + \cdots \\ & = \left (\frac {1}{n + 1} - \frac {1}{(n + 1) (n + 2)}\right ) \\ & \phantom {=} + \left (\frac {1}{(n + 1) (n + 2) (n + 3)} - \frac {1}{(n + 1) (n + 2) (n + 3) (n + 4)}\right ) + \cdots \\ & > \left (\frac {1}{n - 1} - \frac {1}{n + 1}\right ) \\ & \phantom {=} + \left (\frac {1}{(n + 1) (n + 2) (n + 3)} - \frac {1}{(n + 1) (n + 2) (n + 3)}\right ) + \cdots \\ & = 0 + 0 + \cdots \\ & = 0, \end {align*}
using the inequality \[ \frac {1}{(n + 1) \cdots (n + k)} > \frac {1}{(n + 1) \cdots (n + k) (n + k + 1)}. \]
For the right inequality, by grouping consecutive after the first one, we have \begin {align*} g(n) & = \frac {1}{n + 1} - \frac {1}{(n + 1) (n + 2)} + \frac {1}{(n + 1) (n + 2) (n + 3)} \\ & \phantom {=} - \frac {1}{(n + 1) (n + 2) (n + 3) (n + 4)} + \frac {1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5)} - \cdots \\ & = \frac {1}{n + 1} - \left (\frac {1}{(n + 1) (n + 2)} - \frac {1}{(n + 1) (n + 2) (n + 3)}\right ) \\ & \phantom {=} - \left (\frac {1}{(n + 1) (n + 2) (n + 3) (n + 4)} - \frac {1}{(n + 1) (n + 2) (n + 3) (n + 4) (n + 5)}\right ) - \cdots \\ & < \frac {1}{n + 1} - \left (\frac {1}{(n + 1) (n + 2)} - \frac {1}{(n + 1) (n + 2)}\right ) \\ & \phantom {=} - \left (\frac {1}{(n + 1) (n + 2) (n + 3) (n + 4)} - \frac {1}{(n + 1) (n + 2) (n + 3) (n + 4)}\right ) - \cdots \\ & = \frac {1}{n + 1} - 0 - 0 - \cdots \\ & = \frac {1}{n + 1}, \end {align*}
using the inequality \[ \frac {1}{(n + 1) \cdots (n + k - 1) (n + k)} < \frac {1}{(n + 1) \cdots (n + k - 1)}. \]
Hence, \[ 0 < g(n) < \frac {1}{n + 1}. \]
The infinite series for \(e\) is given by \[ e = \sum _{t = 0}^{\infty } \frac {1}{t!}, \] and notice that \[ f(n) = \sum _{t = 1}^{\infty } \frac {n!}{(n + t)!} = n! \sum _{t = 1}^{\infty } \frac {1}{(n + t)!}. \]
Hence, \begin {align*} (2n)! e - f(2n) & = (2n)! \sum _{t = 0}^{\infty } \frac {1}{t!} - (2n)! \sum _{t = 1}^{\infty } \frac {1}{(2n + t)!} \\ & = (2n)! \left (\sum _{t = 0}^{\infty } \frac {1}{t!} - \sum _{t = 2n + 1}^{\infty } \frac {1}{t!}\right ) \\ & = (2n)! \sum _{t = 0}^{2n} \frac {1}{t!} \\ & = \sum _{t = 0}^{2n} \frac {(2n)!}{t!}. \end {align*}
Since \(t \leq 2n\), the terms in the sum represents the number of ways to arrange \((2n - t)\) items out of \(2n\) items, which must be integers. Hence, the sum is an integer as well.
Similarly, the infinite series for \(e^{-1}\) is given by \[ e^{-1} = \sum _{t = 0}^{\infty } \frac {(-1)^t}{t!}, \] and notice that \[ g(n) = - \sum _{t = 1}^{\infty } \frac {(-1)^t n!}{(n + t)!} = - n! \sum _{t = 1}^{\infty } \frac {(-1)^t}{(n + t)!}. \]
Hence, \begin {align*} \frac {(2n)!}{e} + g(2n) & = (2n)! \sum _{t = 0}^{\infty } \frac {(-1)^t}{t!} - (2n)! \sum _{t = 1}^{\infty } \frac {(-1)^t}{(n + t)!} \\ & = (2n)! \left (\sum _{t = 0}^{\infty } \frac {(-1)^t}{t!} - \sum _{t = 2n + 1}^{\infty } \frac {(-1)^t}{t!}\right ) \\ & = (2n)! \sum _{t = 0}^{2n} \frac {(-1)^t}{t!} \\ & = \sum _{t = 0}^{2n} \frac {(-1)^t (2n)!}{t!}, \end {align*}
and by the same argument, since \(t \leq 2n\), this must be an integer as well.
By the previous part, let \(a(n) = f(2n) - (2n)!e\), and \(b(n) = g(2n) + \frac {(2n)!}{e}\), we must have that \(a, b: \NN \to \ZZ \) since they are integers.
Using this notation, \begin {align*} q f(2n) + p g(2n) & = q a(2n) + qe (2n)! + p b(2n) - \frac {p}{e} (2n)! \\ & = q a(2n) + p b(2n) + \left (qe - \frac {p}{e}\right ) (2n)! \\ & = q a(2n) + p b(2n) \end {align*}
must be an integer, since \(p, q, a(2n), b(2n)\) are all integers.
Assume B.W.O.C. that \(e^2\) is irrational. Then there exists natural numbers \(p, q\) such that \[ e^2 = \frac {p}{q} \iff qe = \frac {p}{e}. \]
Since \(e^2 > 1\), \(p > q\).
On one hand, we have \(q f(2n) + p g(2n) > 0\).
On the other hand, let \(n = p\), \begin {align*} q f(2n) + p g(2n) & < q \cdot \frac {1}{2p} + p \cdot \frac {1}{2p + 1} \\ & < q \cdot \frac {1}{2p} + p \cdot \frac {1}{2p} \\ & = \frac {p + q}{2p} \\ & < \frac {2p}{2p} \\ & = 1. \end {align*}
This means \[ 0 < q f(2p) + p g(2p) < 1. \]
But by the previous part, \(q f(2n) + p g(2n)\) is an integer for all positive integer \(n\), and \(n = p\) is a positive integer. This leads to a contradiction.
Hence, such \(p\) and \(q\) does not exist, meaning \(e^2\) is not rational, hence \(e^2\) is irrational.