Selected LaTeX Examples

examples.tex

%% Section 2.1 %%

\section*{Section 2.1}

\subsection*{2.1.2}{Prove that if $\seq{a_n}$ is a convergent sequence, then $\seq{|a_n|}$ is convergent. Is the converse true?}

    \begin{proof} % If $\seq{a_n}$ is convergent, then $\seq{|a_n|}$ is convergent.
        Let $\epsilon>0$ be given. 
        Since $\seq{a_n}$ converges to $a$, there exists $N \in \N$ such that $|a_n - a| < \epsilon, \A n \geq N$. Consider $\seq{|a_n|}$:
        \begin{equation*}
            | |a_n| - a | = | a_n - a | < \epsilon, \A n \geq N
        \end{equation*}
        Thus, it is shown that $|a_n|$ is also a convergent sequence.
    \end{proof}
    
    (Note: The converse is not true.)
    
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\subsection*{2.1.3}{Prove that if the sequence $\seq{|a_n|}$ converges to 0, then $\seq{a_n}$ also converges to 0.}

    \begin{proof}
        Let $\epsilon>0$ be given.
        Since $\seq{|a_n|}$ converges to 0, this means there exists $N \in \N$ such that $||a_n| - 0| < \epsilon, \A n \geq N$.
        \[ ||a_n| - 0| = ||a_n|| = |a_n| = |a_n - 0| < \epsilon\text{, for }n\geq N \]
        Thus, if $|a_n|$ converges to 0, $a_n$ also converges to 0.
    \end{proof}
    
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\subsection*{2.1.6}{Prove that the following limits are 0.}

    \begin{tcolorbox}
        \textbf{Theorem 2.1.6 (Squeeze Theorem)}
        Suppose the sequences $\seq{a_n}$, $\{b_n\}$, and $\{c_n\}$ satisfy 
        $\ds a_n \leq b_n \leq c_n \text{ for all } n \in \N$, 
        and $\linf{n} a_n = \linf{n} c_n = \ell$. Then $\linf{n} b_n = \ell$.
    \end{tcolorbox}
    
    \subsubsection*{(a)}$\ds \linf{n}\frac{n+\cos(n^2-3)}{2n^2+1}$.
    
        \paragraph{Preconditions} Let $a_n, b_n, c_n$ be sequences that satisfy $a_n \leq b_n \leq c_n$, with $a_n \to 0$ and $c_n \to 0$, and \[b_n = \frac{n+\cos(n^2-3)}{2n^2+1}\].
    
        \paragraph{Goal} Use the Squeeze Theorem to show that $a_n \leq b_n \leq c_n$ and that $\linf{n}a_n = \linf{n}c_n = 0$, which leads to $\linf{n}c_n = 0$.
    
        \paragraph{Remarks} I know that $-1 \leq \cos(n^2-3) \leq 1$, so I think that I need to find $a_n < -1$, and $b_n > 1$ for all $n \in \N$, but I am struggling to find $a_n, b_n$ that both meet those conditions and converge to 0, to use for the Squeeze Theorem. 
    
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\subsection*{3.3.12}{Prove that the following limits are 0.}
    
    \subsubsection*{(a)}$\ds \linf{n}\frac{n+\cos(n^2-3)}{2n^2+1}$.

        \paragraph{Preconditions} Let $a_n, b_n, c_n$ be sequences that satisfy $a_n \leq b_n \leq c_n$, with $a_n \to 0$ and $c_n \to 0$, and \[b_n = \frac{n+\cos(n^2-3)}{2n^2+1}\].
    
        \paragraph{Goal} Use the Squeeze Theorem to show that $a_n \leq b_n \leq c_n$ and that $\linf{n}a_n = \linf{n}c_n = 0$, which leads to $\linf{n}c_n = 0$.
    
        \paragraph{Remarks} I know that $-1 \leq \cos(n^2-3) \leq 1$, so I think that I need to find $a_n < -1$, and $b_n > 1$ for all $n \in \N$, but I am struggling to find $a_n, b_n$ that both meet those conditions and converge to 0, to use for the Squeeze Theorem.
        
    \br
    
    \subsubsection*{(b)}$\ds \linf{n}\frac{n+\cos(n^2-3)}{2n^2+1}$.
    
        \begin{equation*}
            | |a_n| - a | = | a_n - a | < \epsilon, \A n \geq N
        \end{equation*}
    
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