\(% 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}}\)

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2015.3.2 Question 2

  1. Let \(m = 1000\). Notice for all \(n \geq m\), \[ n^2 = n \cdot n \geq m \cdot n \geq 1000n. \]

  2. This statement is false. Let \(s_n = (-1)^n\) and \(t_n = -(-1)^n\). Then \(s_n = 1\) and \(t_n = -1\) for even \(n\)s, and \(s_n = -1\) and \(t_n = 1\) for odd \(n\)s.

    So \(s_n \geq t_n\) for even \(n\)s, and \(t_n \geq s_n\) for odd \(n\)s. Since there can be arbitrarily big even and odd numbers, neither of the statements are true for these sequences.

  3. Let \(m_1\) be the \(m\) for \((s_n) \leq (t_n)\) and \(m_2\) be the \(m\) for \((t_n) \leq (u_n)\). Let \(m = \max \{m_1, m_2\}\).

    Notice that for all \(n \geq m\), we have \(n \geq m_1\) and therefore \(s_n leq t_n\), and \(n \geq m_2\) amd therefore \(t_n \leq u_n\).

    By the transitivity of the \(\leq \) relation, we have therefore \(s_n \leq u_n\), for all \(n \geq m\). Therefore, this statement is true.

  4. This statement is true. Let \(m = 4\), we aim to prove that \(2^n \geq n^2\) for all \(n \geq m\).

    We first wish to prove the lemma: for all \(n \geq 4\), we have \(n^2 \geq 2n + 1\).

    This is equivalent to proving that \(n^2 - 2n + 1 \geq 2\) for all \(n \geq 4\).

    Notice that \(n^2 - 2n + 1 = (n - 1)^2 \geq (4 - 1)^2 = 9 \geq 2\) is true.

    This finishes our proof for the lemma.

    We show the original statement by mathematical induction.

    1. Base case. For \(n = 4\), we have \(2^4 = 16 \geq 4^2 = 16\).
    2. Inductive step. Assume that \(2^k \geq k^2\) for some \(n = k \geq 4\). We aim to show that \(2^{k+1} \geq (k+1)^2\). \begin {align*} 2^{k+1} & = 2 \cdot 2^k \\ & \geq 2 \cdot k^2 \\ & = k^2 + k^2 \\ & \geq k^2 + 2k + 1 \\ & = (k + 1)^2. \end {align*}

    Therefore, by the principle of mathematical induction, we have \(2^n \geq n^2\) for all \(n \geq 4\), and this finishes our proof.

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