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\title{Homework 1}
\author{Evgeny Baulin}
\date{11 September 2024}

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    \newpage


    \section{Problem 1}

    \subsection{Problem}
    Using Cauchy’s criterion prove that the series \( \sum_{k=1}^{\infty} \frac{1}{k^3 + 2} \) is convergent.

    \subsection{Solution}

    Let \( \varepsilon > 0 \) be arbitrary, then for \( N\left( \varepsilon \right) = \left[ \frac{1}{\varepsilon} \right] \) and \( \forall m, n \in \mathbb{N}: m \geq W\left( \varepsilon \right) \) we have that:

    \begin{gather*}
        \left| \sum_{k=m+1}^{m+n} \frac{1}{k^3 + 2} \right| = \frac{1}{(m+1)^3 + 2} + \frac{1}{(m+2)^3 + 2} + \ldots + \frac{1}{(m+n)^3 + 2} \leq \frac{1}{(m+1)^3} + \ldots + \frac{1}{(m+n)^3} \leq \\
        \leq \frac{1}{(m+1)^2} + \ldots + \frac{1}{(m+n)^2} \leq \left( \frac{1}{m} - \frac{1}{m + 1} \right) + \ldots + \left( \frac{1}{m + n} - \frac{1}{m + n} \right) = \frac{1}{m} - \frac{1}{m + n} = \frac{1}{m} = \frac{1}{W\left( \varepsilon \right)} = \varepsilon
    \end{gather*}

    Since \( m \geq \mathbb{N}\left( \varepsilon \right) \)

    Therefore, by Cauchy's criterion the series \( \sum_{k=1}^{\infty} \frac{1}{k^3 + 2} \) is convergent


    \section{Problem 2}
    Find the sum of the following series or prove its divergence

    \subsection{Part a}

    \subsubsection{Problem}
    \[
        \sum_{k=2}^{\infty} \frac{3^{k+1}}{6 \cdot 2^{2 k-3}}
    \]

    \subsubsection{Solution}

    \[
        \frac{3^{k+1}}{6 \cdot 2^{2k-3}} = \frac{3^k}{\frac{1}{4} \cdot 2^{2k}} = 4 \cdot \frac{3^k}{4^k} = 4 \cdot \left( \frac{3}{4} \right)^k \Rightarrow \sum_{k=2}^{\infty} \left( \frac{3}{4} \right)^k
    \]

    This series is convergent since \( \left| L = \frac{3}{4} \right| < 1 \) and the sum is equal to \( 4\frac{\frac{3}{4}}{1 - \frac{3}{4} - \frac{3}{4}} = 9 \)

    Answer: \(9\)

    \subsection{Part b}

    \subsubsection{Problem}
    \[
        \sum_{k=1}^{\infty}(\sqrt{k+2}-2 \sqrt{k+1}+\sqrt{k})
    \]

    \subsubsection{Solution}

    \begin{gather*}
        \sum_{k=1}^{\infty}(\sqrt{k+2}-\sqrt{k+1}) - \sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k}) \\
        S_n = \sum_{k=1}^{\infty}(\sqrt{k+2}-\sqrt{k+1}) - \sum_{k=1}^{\infty}(\sqrt{k+1}-\sqrt{k}) = -\sqrt{2} - \sqrt{n + \infty} + \sqrt {1} - \sqrt {n + 1} = \\
        = \sqrt{n + 2} - \sqrt{n + 1} + \sqrt{1} - \sqrt{2}\\
        \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sqrt{n + 2} - \sqrt{n + 1} + \sqrt{1} - \sqrt{2} = \infty - \infty + 1 - \sqrt{2} \text{ so limit does not exist} \Rightarrow \text{series is divergent}
    \end{gather*}

    \subsection{Part c}

    \subsubsection{Problem}
    \[
        \sum_{k=1}^{\infty}\left(\frac{k^3}{k^3+2}\right)^{k^3}
    \]

    \subsubsection{Solution}

    \subsection{Part d}

    \subsubsection{Problem}
    \[
        \sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^2}\right)
    \]

    \subsubsection{Solution}

    \subsection{Part e}

    \subsubsection{Problem}
    \[
        \sum_{k=1}^{\infty} e^{-k}\left(1+\frac{1}{k}\right)^{k^2}
    \]

    \subsubsection{Solution}

    \subsection{Part f}

    \subsubsection{Problem}
    \[
        \sum_{k=1}^{\infty}\left(\frac{1}{k^2+4 k+3}\right)
    \]

    \subsubsection{Solution}


\end{document}