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Week 6 HW

**19.2 Ex 2:** We prove \(\frac{n+1}{n!} < \frac{8}{2^n}\) by induction.

Base Case: \(\frac{1+1}{1!} < \frac{8}{2}\)

General Case: We assume \(\frac{n+1}{n!} < \frac{8}{2^n}\) and we must prove \(\frac{n+2}{(n+1)!} < \frac{8}{2^{n+1}}\) which is true if and only if \(\frac{n+2}{n!(n+1)} < \frac{8}{2^n}\frac{1}{2} \Longleftrightarrow \frac{2n+4}{n!(n+1)} < \frac{8}{2^n} \Longleftrightarrow \)

\(\frac{2n+4}{n!(n+1)} < \frac{n+1}{n!} \Longleftrightarrow \frac{2n+4}{n+1} < {n+1} \Longleftrightarrow 2n+4 < n^2+2n+1 \Longleftrightarrow 3 < n^2\) which is true for n>1.

It follows that \(\sum{}{}{\frac{n+1}{n!}}\) is convergent because each of its terms is smaller than each term of a multiple of a convergent \(p\)-series.

**19.2 Ex 4:** Each term of series (1) is less than each term of series (2) so since series (2) is a convergent p-series (1) is convergent.

**19.2 Ex 6:** If \(\frac{a_n}{b_n} \Rightarrow 0\) then \(\frac{a_n}{b_n} < 1\) for sufficiently large n so \(a_n < b_n\) after a certain point so if \(b_n\) is convergent so is \(a_n\). Similarly If \(\frac{a_n}{b_n} \Rightarrow \infty\) then \(\frac{a_n}{b_n} > 1\) for sufficiently large n so \(a_n > b_n\) after a certain point so if \(b_n\) is divergent so is \(a_n\).

**19.2 Ex 8:** \((\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3})+ \cdot\cdot\cdot - \frac{1}{n+1} = 1 - \frac{1}{n+1}\) which goes to 1 as \(n\) becomes large.

**19.3 Ex 1:** \(0 \le |x^n| \le 1\) when \(|x| \le 1\) so \(|c_nx^n| = c_n|x^n| \le c_n\). Therefore since \(\sum{}{}{c_n}\) is convergent, \(\sum{}{}{c_nx^n}\) is absolutely convergent (by the comparison test).

**19.3 Ex 2:** \(0 \le |\sin n\theta| \le 1\) for all theta so \(|r^n\sin n\theta| = r^n|\sin n\theta| \le r^n\). Therefore since \(\sum{}{}{r^n}\) is convergent, \(\sum{}{}{r^n\sin n\theta}\) is absolutely convergent (by the comparison test). The proof for \(\cos\) is the same.

**19.3 Ex 3:**

a) Yes. The terms of \(1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + ... + \frac{1}{(2n+1)^2}\) (with appropriate zeros inserted) are less than the terms of a convergent \(p\)-series, so it is convergent by the comparison test.

b) No. If the series was absolutely convergent both the series of positive and negative terms would be convergent, but the series of negative terms is a divergent \(p\)-series.

c) No. The series \(\sum{}{}{\frac{1}{(n-1)\log{n}}}\) is divergent by the integral and comparison tests.

d) Yes. The absolute value of each term is \(\le \frac{1}{n^2}\) which is the \(n\)-th term of a convergent \(p\)-series.

**19.32 Ex 1:**

a) This series is alternating positive and negative terms. \[\frac{n + 1}{(n+2)^2} \le \frac{n}{(n+1)^2} \Longleftrightarrow \frac{1}{n^3 + 4n^4 + 4n} \le \frac{1}{n^3 + 3n^2 + 3n + 1} \Longleftrightarrow n^3 + 3n^2 + 3n + 1 \le n^3 + 4n^2 + 4n \Longleftrightarrow 1 \le n^2 + n\] so the absolute value of each term of the series is decreasing. \[\lim_{n\to\infty}\frac{n}{(n+1)^2} = 0\] Therefore the series is convergent.

b) This series is alternating positive and negative terms. \[\ln\frac{n+2}{n+1} \le \ln\frac{n+1}{n} \Longleftrightarrow \frac{n+2}{n+1} \le \frac{n+1}{n} \Longleftrightarrow n^2 + 2n \le n^2 + 2n + 1\] so the absolute value of each term of the series is decreasing. \[\lim_{n\to\infty}\ln\frac{n+1}{n} = \lim_{n\to\infty}\ln\frac{1+\frac{1}{n}}{1} = \ln1 = 0\] Therefore the series is convergent.

c) This series is alternating positive and negative terms. \[\frac{\sqrt{n+1}}{2+n} \le \frac{\sqrt{n}}{1+n} \Longleftrightarrow \frac{n+1}{(2+n)^2} \le \frac{n}{(1+n)^2} \Longleftrightarrow n^3 + 3n^2 + 3n + 1 \le n^3 + 4n^2 + 4n\] so the absolute value of each term of the series is decreasing. \[\lim_{n\to\infty}\frac{\sqrt{n}}{1+n} = \lim_{n\to\infty}\frac{1}{\frac{1}{\sqrt{n}}+\sqrt{n}} = 0\] Therefore the series is convergent.

d) This series is alternating positive and negative terms. \[\frac{1\cdot3\cdot\cdot\cdot(2n-1)\cdot(2n+1)}{2\cdot4\cdot\cdot\cdot2n\cdot(2n+2)\cdot(2n+4)} \le\]

\[\frac{1\cdot3\cdot\cdot\cdot(2n-1)}{2\cdot4\cdot\cdot\cdot2n\cdot(2n+2)} \Longleftrightarrow 2n+1\le2n+4\] so the absolute value of each term of the series is decreasing. \[\lim_{n\to\infty}\frac{1\cdot3\cdot\cdot\cdot(2n-1)}{2\cdot4\cdot\cdot\cdot2n\cdot(2n+2)} = \lim_{n\to\infty}\frac{1}{2} \cdot \frac{3}{4} \cdot\cdot\cdot \frac{2n-1}{2n} \cdot \frac{1}{2n+2}\]

\[< \lim_{n\to\infty}1\cdot1\cdot\cdot\cdot1\cdot \frac{1}{2n+2} = \lim_{n\to\infty}\frac{1}{2n+2} = 0\] Therefore the series is convergent.

e) This series is alternating positive and negative terms. \[\frac{3\cdot6\cdot\cdot\cdot3n\cdot(3n+3)}{1\cdot4\cdot\cdot\cdot(3n-2) \cdot(3n+1)} \cdot \frac{1}{(n+1)^2} \le \frac{3\cdot6\cdot\cdot\cdot3n}{1\cdot4\cdot\cdot\cdot(3n-2)} \cdot \frac{1}{n^2}\Longleftrightarrow \frac{3n+3}{3n+1} \le \frac{(n=1)^2}{n^2} \Longleftrightarrow 3n^3+3n^2 \le 3n^3 + 7n^2 + 5n + 1\] so the absolute value of each term of the series is decreasing. \[\lim_{n\to\infty}\frac{3\cdot6\cdot\cdot\cdot3n}{1\cdot4\cdot\cdot\cdot(3n-2)} \cdot \frac{1}{n^2} = \lim_{n\to\infty}3\cdot \frac{1}{1} \cdot \frac{3}{4}\cdot \frac{6}{7}\cdot\cdot\cdot\frac{3n-3}{3n-2}\cdot\frac{1}{n} \le 3 \lim_{n\to\infty} 1\cdot1\cdot1\cdot\cdot\cdot\frac{1}{n} = 0\] Therefore the series is convergent.

**19.32 Ex 2:**

a) The series is alternating positive and negative terms. \[\frac{\ln(n+1)}{\sqrt{n+1}} \le \frac{\ln n}{\sqrt{n}}\Longleftrightarrow \frac{d}{d x} \frac{\ln x}{\sqrt x} = \frac{2-\ln x}{2x\sqrt{x}}\le 0\] which is true for all sufficiently large x, so the absolute value of each term of the series is decreasing. \[\lim_{n\to\infty}\frac{\ln n}{\sqrt{n}} = \lim_{n\to\infty}-\frac{1}{2n^\frac{5}{2}}= 0\] So the series is convergent.

b) The series is alternating positive and negative terms.\[(\frac{n+1}{n+2})^{n+1} \le (\frac{n}{n+1})^{n} \Longleftrightarrow (1-\frac{1}{n+2})(1-\frac{1}{n+2})^{n} \le (1-\frac{1}{n+1})^{n}\] so the absolute value of each term of the series is decreasing.

By l’Hospital’s rule: \[\lim_{n\to\infty}(1-\frac{1}{n+1})^n = e^{\lim_{n\to\infty}n\ln\frac{n}{n+1}} = e^{\lim_{n\to\infty}\frac{\ln\frac{n}{n+1}}{\frac{1}{n}}} = e^{-\lim_{n\to\infty}\frac{n^3+n^2}{n}} = e^{-\infty}= 0\] so the series converges.

c) Using l’Hospital’s rule we see that: \[\lim_{n\to\infty}1-n\ln\frac{n+1}{n} = 1 - \lim_{n\to\infty}\frac{\frac{n}{n+1}}{-\frac{1}{n^2}} = 1+ \lim_{n\to\infty}\frac{n^3}{n+1} = \infty \ne 0\] so the series diverges.

d) This series is \(-\frac{1}{3}\) of the harmonic series so the series diverges.

**19.32 Ex 3:**

The series alternates positive and negative terms. \[\frac{d}{d x} \frac{(\ln n)^p}{n^q} = \frac{p(\ln n)^{p-1}}{n^{q+1}} - \frac{q (\ln n)^p}{n^{q+1}} \le 0 \Longleftrightarrow p \le q \ln n\] which is true for sufficiently large n. Using l’Hospital rule \(\lceil p \rceil\) times gives us that \[\lim_{n\to\infty}\frac{(\ln n)^p}{n^q} = 0\]

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