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11.8 Cauchy sequences

TutorialsWorshopsHomework 9.pdf

  1. Compute the following limit \(\lim_{n\to\infty}\left(\frac{2n+3}{2n}\right)^{3n+2}=\lim_{n\to\infty}\left(1+\frac{3}{2n}\right)^{3n}\cdot\lim_{n\to\infty}\left(1+\frac{3}{2n}\right)^2=\lim_{n\to\infty}\left(1+\frac{3}{2n}\right)^{3n}=\left(\lim_{n\to\infty}\left(1+\frac{3}{2n}\right)^{2n}\right)^{\frac32}=\left(\lim_{n\to\infty}\left(1+\frac{3}{n}\right)^{n}\right)^{\frac32}=\left(e^3\right)^{\frac32}=e^{\frac92}\)

  2. Let \((x_n)\) and \((y_n)\) be Cauchy sequences. Decide whether each of the following sequences is a Cauchy sequence, justifying each conclusion.

    (a) \(z_n = |x_n - y_n|\)

    Since \((x_n)\) and \((y_n)\) are Cauchy sequences, then

    \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|x_{n_1}-x_{n_2}|<\frac{\varepsilon}{2}\)

    \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|y_{n_1}-y_{n_2}|<\frac{\varepsilon}{2}\)

    We want \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow\left||x_{n_1}-y_{n_1}|-|x_{n_2}-y_{n_2}|\right|<\varepsilon\)

    Since we know \(\left||x_{n_1}-y_{n_1}|-|x_{n_2}-y_{n_2}|\right|\leq|x_{n_1}-y_{n_1}-x_{n_2}+y_{n_2}|\leq|x_{n_1}-x_{n_2}\left|+\right|y_{n_1}-y_{n_2}|\), then N.T.P. \(|x_{n_1}-y_{n_1}\left|+\right|x_{n_2}-y_{n_2}|<\varepsilon\)

    Obviously...

    (b) \(z_n = (-1)^n x_n\)

    Since \((x_n)\) is Cauchy sequences, then \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|x_{n_1}-x_{n_2}|<\varepsilon\)

    We want \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|\left(-1\right)^{n_1}x_{n_1}-\left(-1\right)^{n_2}x_{n_2}|<\varepsilon\)

    We consider two cases

    If \(n_1,n_2\) are both even or odd, then \(|x_{n_1}-x_{n_2}|<\varepsilon\), we are done

    If \(n_1,n_2\) are either even or odd, then \(|\left(-1\right)^{n_1}x_{n_1}-\left(-1\right)^{n_2}x_{n_2}|=|x_{n_1}+x_{n_2}|\)​

    Then i don't know how to do with it.

    After trying some examples, I know that it is not a convergent sequence.

    Thus it is not a Cauchy Sequence.

    Since \(x_n\) is a Cauchy sequence, thus \(\lim_{n\to\infty}x_{n}=L\)​

    If \(L=0\), then \(\lim_{n\to\infty}x_{n}=0=\lim_{n\to\infty}-x_{n}=\lim_{n\to\infty}\left(-1\right)^{n}x_{n}\)​

    If \(L\neq 0\), then \(\lim_{n\to\infty}x_{n}=L,\lim_{n\to\infty}-x_{n}=-L,\lim_{n\to\infty}\left(-1\right)^{n}x_{n}\) is not a fixed value.

    Thus it is not convergent, and \(z_n\) is not a Cauchy sequence

  3. Decide if the following series converges:

    (a) \(\sum_{n=1}^{\infty} \frac{(-1)^n n}{n + 1}\) diverge

    (b) \(\sum_{n=5}^{\infty} \frac{2^n - 5^n}{9^n}\) converge

    (c) (telescope sum)\(\sum_{n=1}^{\infty} \frac{1}{(2n - 1)(2n + 1)}\) converge

    (d) \(\sum_{n=1}^{\infty}\sqrt[n]{n}\) diverge

    (e) \(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^4}\) converge

    (f) \(\sum_{n=1}^{\infty}\pi^{n/2}\cos(n\pi)\) diverge

    (g) \(\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n^2+1}}\)

    Since \(\frac{1}{\sqrt[3]{n^2+1}}\geq\frac{1}{\sqrt[3]{2n^2}}=\frac{1}{\sqrt[3]{2}}\cdot\frac{1}{\sqrt[3]{n^2}}\) and \(\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n^2}}\) is divergent, thus \(\sum_{n=1}^{\infty}\frac{1}{\sqrt[3]{n^2+1}}\) is divergent

    (h) \(\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2}\) converge

    (i) \(\sum_{n=1}^{\infty} \frac{\left(1 + \frac{1}{n}\right)^n}{2^n}\) converge

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