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5

Homework 5.pdf

Homework 5

Week 5 November 16, 2024

1. (20 points) Prove that \(\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}\) converges conditionally. Hint: Use the sequence of partial sums.

Proof

Let's \(a_n=(-1)^{n}\frac{1}{\sqrt{n}}\), then \(\sum_{n=1}^{\infty}a_{n}=-1+\frac{1}{\sqrt2}-\frac{1}{\sqrt3}+\frac{1}{\sqrt4}+\cdots\)

Let use partial sum, consider a new sequence \(b_n\)

\(b_1=\frac{1}{\sqrt2}-\frac{1}{\sqrt3},b_2=\frac{1}{\sqrt4}-\frac{1}{\sqrt5},\ldots,b_{n}=\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}},\ldots\)

Then \(\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}b_{n}-1\)

Since \(b_{n}=\frac{1}{\sqrt{2n}}-\frac{1}{\sqrt{2n+1}}=\frac{\sqrt{2n+1}-\sqrt{2n}}{\sqrt{2n}\cdot\sqrt{2n+1}}=\frac{\left(\sqrt{2n+1}-\sqrt{2n}\right)\left(\sqrt{2n+1}+\sqrt{2n}\right)}{\sqrt{2n}\cdot\sqrt{2n+1}(\sqrt{2n+1}+\sqrt{2n})}=\frac{1}{\sqrt{2n}\cdot\sqrt{2n+1}(\sqrt{2n+1}+\sqrt{2n})}\)

\(0<b_{n}=\frac{1}{\sqrt{2n}\cdot\sqrt{2n+1}(\sqrt{2n+1}+\sqrt{2n})}\leq\frac{1}{2n\cdot2\sqrt{2n}}=\frac{1}{4\sqrt2n^{\frac32}}\)

Thus since \(\sum\frac{1}{4\sqrt2n^{\frac32}}\) is convergent(\(p=\frac32>1\)), thus by theorem \(\sum b_n\) is convergent, then \(\sum b_{n}-1\) is convergent

Finally, \(\sum a_n\) is convergent.

Then we need to prove it is conditionally convergent

Take \(\sum_{n=1}^{\infty}\left|(-1)^{n}\frac{1}{\sqrt{n}}\right|=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}\), since \(p=\frac12<1\), then \(\sum_{n=1}^{\infty}\left|(-1)^{n}\frac{1}{\sqrt{n}}\right|\) is divergent

Since \(\sum_{n=1}^{\infty}\left|(-1)^{n}\frac{1}{\sqrt{n}}\right|\) is divergent and \(\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{\sqrt{n}}\) is convergent.

Then \(\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{\sqrt{n}}\) is conditionally convergent.

2. Consider each of the following propositions. Provide short proofs for those that are true and counterexamples for those that are not.

  1. (8 points) If \(\sum a_n\) converges absolutely, then \(\sum a_{2n}\) also converges absolutely.

    True

    Since \(a_{2n}\) is the subsequence of \(a_n\), then \(0\leq|a_{2n}|\leq|a_{n}|\)​

    Since \(\sum |a_n|\) converges, then by comparing test \(\sum|a_{2n}|\) converges

  2. (8 points) If \(\sum a_n\) converges and \((b_n)\) converges, then \(\sum a_n b_n\) converges.

    False

    Take \(a_{n}=b_{n}=\frac{(-1)^{n}}{\sqrt{n}}\), then \(\sum a_n\) converges and \((b_n)\) converges

    But \(\sum a_{n}b_{n}=\sum\frac{1}{n}\) is divergent.

  3. (9 points) If \(\sum a_n\) converges conditionally, then \(\sum n^2 a_n\) diverges.

    True

    Suppose \(\sum n^2 a_n\) is convergent, we know \(\sum\left|a_{n}\right|\) diverges

    Then by theorem \(n^2a_n\to 0\), thus \(n^2|a_n|\to 0\)​

    Therefore \(|a_{n}|\leq\frac{1}{n^2}\) for some \(n\) \(\Rightarrow\) \(\sum|a_{n}|\leq\sum\frac{1}{n^2}\)​

    Since \(\sum\frac{1}{n^2}\) is convergent, by theorem \(\sum|a_{n}|\) is also convergent

    Contradiction!

    Thus \(\sum n^2 a_n\) diverges.

3. Decide if the following series converge or diverge.

  1. (10 points) \(\sum \frac{n(n+1)}{4^n}\)

    Use D'Alembert Criterion

    \(\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_{n}|}=\lim_{n\to\infty}\frac{\left(n+1\right)\left(n+2\right)4^{n}}{n\left(n+1\right)4^{n+1}}=\lim_{n\to\infty}\frac{\left(n+2\right)}{4n}=\frac14<1\)​

    Thus \(\sum \frac{n(n+1)}{4^n}\) is convergent

  2. (10 points) \(\sum \frac{1}{n^{1 + \frac{1}{n}}}\)

    Consider \(a_n=\frac{1}{n^{1+\frac{1}{n}}}=\frac{1}{n\cdot\sqrt[n]{n}}\approx\frac{1}{n}=b_n\), then \(\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=\lim_{n\to\infty}\frac{\frac{1}{n^{1+\frac{1}{n}}}}{\frac{1}{n}}=\lim_{n\to\infty}\frac{n}{n\cdot\sqrt[n]{n}}=1\)

    Since \(\sum \frac1n\) is divergent, thus \(\sum \frac{1}{n^{1 + \frac{1}{n}}}\) is divergent

  3. (10 points) \(\sum \frac{n + 4^n}{n + 6^n}\)

    Consider \(\frac{n+4^{n}}{n+6^{n}}\leq\frac{n+4^{n}}{6^{n}}\leq2\frac{4^{n}}{6^{n}}=2\cdot\left(\frac23\right)^{n}\)

    Since \(\sum 2\cdot (\frac23)^n\) is convergent, thus \(\sum \frac{n + 4^n}{n + 6^n}\) is convergent

4. (25 points) Let \(B = \{x \in \mathbb{Q} : 0 < x < 1\}\).

  1. What are the limit points of \(B\)? Justify.

    When \(0<x_0<1\), it satisfies \(\forall\delta>0:\left\lbrack(x_0-\delta,x_0)\cup(x_0,x_0+\delta)\right\rbrack\cap X\neq\emptyset\) Yes

    Since rational number is dense in real number, there always exists rational numbers in the interval

    When \(x_0=0\), it satisfies \(\forall\delta>0:(x_0,x_0+\delta)\cap X\neq\emptyset\) Yes

    When \(x_0=1\), it satisfies \(\forall\delta>0:(x_0-\delta,x_0)\cap X\neq\emptyset\) Yes

    When \(x_0<0\cup x_0>1\), there is no limit points No

    Thus the limit points of \(B\) are \([0,1]\)​

  2. Describe the inner set of \(B\).

    \(\empty\)

  3. What is the boundary of \(B\)? Justify.

    We use equivalence of definition: \(\forall \delta > 0 \quad (x_0 - \delta, x_0 + \delta) \cap A \neq \emptyset\) and \((x_0 - \delta, x_0 + \delta) \cap A^c \neq \emptyset\)

    When \(0<x_0<1\), Since rational number is dense in real number, there always exists rational numbers in the interval s.t. \((x_0-\delta,x_0+\delta)\cap A\neq\emptyset\)

    Also, since the complement of \(B\) is the real number in \((-\infty,0]\cup\left\lbrack1,+\infty\right)\) and irrational number in \(B\)

    Thus \((x_0 - \delta, x_0 + \delta) \cap A^c \neq \emptyset\) is also true

    When \(x_0=0\) or \(1\), it satisfies \(\forall \delta > 0 \quad (x_0 - \delta, x_0 + \delta) \cap A \neq \emptyset\) and \((x_0 - \delta, x_0 + \delta) \cap A^c \neq \emptyset\) Yes

    When \(x_0<0\), \(\exists\delta>0\) such that \(x_0-\delta<0,x_0+\delta<0\), then \((x_0-\delta,x_0+\delta)\cap A=\emptyset\) No

    When \(x_0>1\), \(\exists\delta>0\) such that \(x_0-\delta>1,x_0+\delta>1\), then \((x_0-\delta,x_0+\delta)\cap A=\emptyset\) No

    Thus the boundary of \(B\) is \([0,1]\)

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