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Schaum's outline of theory and problems of college mathematics : algebra, discrete mathematics, precalculus, introduction to caculus

Frank Ayres; Philip A Schmidt

Chapter 50

Infinite Series - all with Video Answers

Educators

Chapter Questions

Problem 1

Examine the infinite geometric series $a+a r+a r^2+\cdots+a r^n+\cdots$ for convergence and divergence.
The sum of the first $n$ terms is $S_n=\frac{a\left(1-r^n\right)}{1-r}=\frac{a}{1-r}\left(1-r^n\right)$.
If $|r|<1, \lim _{n \rightarrow \infty} r^n=0$; then $\lim _{n \rightarrow \infty} S_n=\frac{a}{1-r}$ and the series is convergent.
If $|r|>1, \lim _{n \rightarrow \infty} r^n$ does not exist and $\lim _{n \rightarrow \infty} S_n$ does not exist; the series is divergent.
If $r=1$, the series is $a+a+a+\cdots+a+\cdots ;$ then $\lim _{n \rightarrow \infty} S_n=\lim _{n \rightarrow \infty} n a$ does not exist.
If $r=-1$, the series is $a-a+a-a+\cdots ;$ then $S_n=a$ or 0 according as $n$ is odd or even, and $\lim _{n \rightarrow \infty} S_n$ does not exist.
Thus, the infinite geometric series convergence to $\frac{a}{1-r}$ when $|r|<1$, and diverges when $|r| \geq 1$.

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01:39

Problem 2

Show that the following series are convergent:
(a) $1+\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots$.
This is a geometric series with ratio $r=\frac{2}{3}$; then $|r|<1$ and the series is convergent.
(b) $2-\frac{3}{2}+\frac{9}{8}-\frac{27}{32}+\cdots+(-1)^{n-1} 2\left(\frac{3}{4}\right)^{n-1}+\cdots$.
This is a geometric series with ratio $r=-\frac{3}{4}$; then $|r|<1$ and the series is convergent.
(c) $1+\frac{1}{2^p}+\frac{1}{2^p}+\frac{1}{4^p}+\frac{1}{4^p}+\frac{1}{4^p}+\frac{1}{4^p}+\frac{1}{8^p}+\cdots, p>1$.
This series may be rewritten as $1+\frac{2}{2^p}+\frac{4}{4^p}+\frac{8}{8^p}+\cdots$, a geometric series with ratio $r=2 / 2^p$. Since $|r|<1$, when $p>1$, the series is convergent.

Wendi Zhao
Wendi Zhao
Numerade Educator
01:03

Problem 3

Show that the following series are divergent:
(a) $2+\frac{3}{2}+\frac{4}{3}+\cdots+\frac{n+1}{n}+\cdots$.
Since $\lim _{n \rightarrow \infty} s_n=\lim _{n \rightarrow \infty} \frac{n+1}{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=1 \neq 0$, the series is divergent.
(b) $\frac{1}{2}+\frac{3}{8}+\frac{5}{16}+\frac{9}{32}+\cdots+\frac{2^{n-1}+1}{4 \cdot 2^{n-1}}+\cdots$.
Since $\lim _{n \rightarrow \infty} s_n=\lim _{n \rightarrow \infty}+\frac{2^{n-1}+1}{4 \cdot 2^{n-1}}=\lim _{n \rightarrow \infty} \frac{1+1 / 2^{n-1}}{4}=\frac{1}{4} \neq 0$, the series is divergent.

Tyler Moulton
Tyler Moulton
Numerade Educator
01:03

Problem 4

Show that $1+\frac{1}{2^p}+\frac{1}{3^p}+\frac{1}{4^p}+\cdots+\frac{1}{n^p}+\cdots$ is divergent for $p \leq 1$ and convergent for $p>1$.
For $p=1$, the series is the harmonic series and is divergent.
For $p<1$, including negative values, $\frac{1}{n^p} \geq \frac{1}{n}$, for every $n$. Since every term of the given series is equal to or greater than the corresponding term of the harmonic series, the given series is divergent. For $p>1$, compare the series with the convergent series
$$
1+\frac{1}{2^p}+\frac{1}{2^p}+\frac{1}{4^p}+\frac{1}{4^p}+\frac{1}{4^p}+\frac{1}{4^p}+\frac{1}{8^p}+\cdots
$$
of Problem $50.2(\mathrm{c})$. Since each term of the given series is less than or equal to the corresponding term of series (1), the given series is convergent.

Tyler Moulton
Tyler Moulton
Numerade Educator
00:31

Problem 5

Use Problem 50.4 to determine whether the following series are convergent or divergent:
(a) $1+\frac{1}{2 \sqrt{2}}+\frac{1}{3 \sqrt{3}}+\frac{1}{4 \sqrt{4}}+\cdots$. The general term is $\frac{1}{n \sqrt{n}}=\frac{1}{n^{3 / 2}}$.
This is a $p$ series with $p=\frac{3}{2}>1$; the series is convergent.
(b) $1+4+9+16+\cdots$. The general term is $n^2=\frac{1}{n^{-2}}$.
This is a $p$ series with $p=-2<1$; the series is divergent.
(c) $1+\frac{\sqrt[4]{2}}{4}+\frac{\sqrt[4]{3}}{9}+\frac{\sqrt[4]{4}}{16}+\cdots$. The general term is $\frac{\sqrt[4]{n}}{n^2}=\frac{1}{n^{7 / 4}}$.
The series is convergent since $p=\frac{7}{4}>1$.

Adrian Co
Adrian Co
Numerade Educator
01:08

Problem 6

Use the comparison test to determine whether each of the following is convergent or divergent:
(a) $1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\cdots$
The general term $\frac{1}{n !} \leq \frac{1}{n^2}$. Thus, the terms of the given series are less than or equal to the corresponding terms of the $p$ series with $p=2$. The series is convergent.
(b) $\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{9}+\cdots$.
The general term $\frac{1}{1+2^{n-1}} \leq \frac{1}{2^{n-1}}$. Thus, the terms of the given series are less than or equal to the corresponding terms of the geometric series with $a=1$ and $r=\frac{1}{2}$. The series is convergent.
(c) $\frac{2}{1}+\frac{3}{4}+\frac{4}{9}+\frac{5}{16}+\cdots$.
The general term $\frac{n+1}{n^2}=\frac{1}{n}+\frac{1}{n^2}>\frac{1}{n}$. Thus, the terms of the given series are equal to or greater than the corresponding terms of the harmonic series. The series is divergent.
(d) $\frac{1}{3}+\frac{1}{12}+\frac{1}{27}+\frac{1}{48}+\cdots$.
The general term $\frac{1}{3 \cdot n^2} \leq \frac{1}{n^2}$. Thus, the terms of the given series are less than or equal to the corresponding terms of the $p$ series with $p=2$. The series is convergent.
(e) $1+\frac{1}{2}+\frac{1}{3^2}+\frac{1}{4^3}+\frac{1}{5^4}+\cdots$
The general term $\frac{1}{n^{n-1}} \leq \frac{1}{n^2}$ for $n \geq 3$. Thus, neglecting the first two terms, the given series is term by term less than or equal to the corresponding terms of the $p$ series with $p=2$. The given series is convergent.

Nick Johnson
Nick Johnson
Numerade Educator

Problem 7

Apply the ratio test to each of the following. If it fails, use some other method to determine convergency or divergency.
(a) $\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+\cdots \quad$ or $\quad \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+\cdots$
For this series $s_n=\frac{n}{2^n}, s_{n+1}=\frac{n+1}{2^{n+1}}$, and $r_n=\frac{s_{n+1}}{s_n}=\frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n}=\frac{n+1}{2 n}$. Then $R=\lim _{n \rightarrow \infty} r_n=$ $\lim _{n \rightarrow \infty} \frac{n+1}{2 n}=\lim _{n \rightarrow \infty} \frac{1+1 / n}{2}=\frac{1}{2}<1$ and the series is convergent.
(b) $3+\frac{9}{2}+\frac{9}{2}+\frac{27}{8}+\frac{81}{40}+\cdots \quad$ or $\quad \frac{3}{1 !}+\frac{3^2}{2 !}+\frac{3^3}{3 !}+\cdots$.
Here $s_n=\frac{3^n}{n !}, s_{n+1}=\frac{3^{n+1}}{(n+1) !}$, and $r_n=\frac{3^{n+1}}{(n+1) !} \cdot \frac{n !}{3^n}=\frac{3}{n+1}$. Then $R=\lim _{n \rightarrow \infty} \frac{3}{n+1}=0$ and the series is convergent.
(c) $\frac{1}{1 \cdot 1}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{4 \cdot 7}+\cdots$.
Here $s_n=\frac{1}{n(2 n-1)}, s_{n+1}=\frac{1}{(n+1)(2 n+1)}$, and $r_n=\frac{n(2 n-1)}{(n+1)(2 n+1)}$.
Then $R=\lim _{n \rightarrow \infty} \frac{n(2 n-1)}{(n+1)(2 n+1)}=\lim _{n \rightarrow \infty} \frac{2-1 / n}{(1+1 / n)(2+1 / n)}=1$ and the test fails.
Since $\frac{1}{n(2 n-1)} \leq \frac{1}{n^2}$, the given series is term by term less than or equal to the convergent $p$ series, with $p=2$. The given series is convergent.
(d) $\frac{2}{1^2+1}+\frac{2^3}{2^2+2}+\frac{2^5}{3^2+3}+\cdots$.
Here $s_n=\frac{2^{2 n-1}}{n^2+n}, s_{n+1}=\frac{2^{2 n+1}}{(n+1)^2+(n+1)}$, and $r_n=\frac{2^{2 n+1}}{(n+1)(n+2)} \cdot \frac{n(n+1)}{2^{2 n-1}}=\frac{4 n}{n+2}$. Then $R=$ $\lim _{n \rightarrow \infty} \frac{4}{1+2 / n}=4$ and the series is divergent.
(e) $\frac{1}{5}+\frac{2}{25}+\frac{6}{125}+\frac{24}{625}+\cdots$.
In this series $s_n=\frac{n !}{5^n}, s_{n+1}=\frac{(n+1) !}{5^{n+1}}$, and $r_n=\frac{(n+1) !}{5^{n+1}} \cdot \frac{5^n}{n !}=\frac{n+1}{5}$. Now $\lim _{n \rightarrow \infty} r_n$ does not exist. However, since $s_n \rightarrow \infty$ as $n \rightarrow \infty$, the series is divergent.

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02:19

Problem 8

Test the following alternating series for convergence:
(a) $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$
$s_n>s_{n+1}$, for all values of $n$, and $\lim _{n \rightarrow \infty} s_n=\lim _{n \rightarrow \infty} \frac{1}{2 n-1}=0$. The series is convergent.
(b) $\frac{1}{2^3}-\frac{2}{3^3}+\frac{3}{4^3}-\frac{4}{5^3}+\cdots$.
$s_n>s_{n+1}$, for all values of $n$, and $\lim _{n \rightarrow \infty} s_n=\lim _{n \rightarrow \infty} \frac{n}{(n+1)^3}=0$. The series is convergent.

Kumar Vaibhav
Kumar Vaibhav
Numerade Educator

Problem 9

Investigate the following for absolute convergence, conditional convergence, or divergence:
(a) $1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots$.
Here $\left|s_n\right|=\frac{1}{2^{n-1}},\left|s_{n+1}\right|=\frac{1}{2^n}$, and $R=\lim _{n \rightarrow \infty} \frac{\left|s_{n+1}\right|}{\left|s_n\right|}=\lim _{n \rightarrow \infty} \frac{2^{n-1}}{2^n}=\frac{1}{2}<1$. The series is absolutely convergent.
(b) $1-\frac{4}{1 !}+\frac{4^2}{2 !}-\frac{4^3}{3 !}+\cdots$
Here $\left|s_n\right|=\frac{4^{n-1}}{(n-1) !},\left|s_{n+1}\right|=\frac{4^n}{n !}$, and $R=\lim _{n \rightarrow \infty} \frac{4}{n}=0$. The series is absolutely convergent.
(c) $\frac{1}{2-\sqrt{2}}-\frac{1}{3-\sqrt{3}}+\frac{1}{4-\sqrt{4}}-\frac{1}{5-\sqrt{5}}+\cdots$.
The ratio test fails here.
Since $\frac{1}{n+1-\sqrt{n+1}}>\frac{1}{n+2-\sqrt{n+2}}$ and $\lim _{n \rightarrow \infty} \frac{1}{n+1-\sqrt{n+1}}=0$, the series is convergent.
Since $\frac{1}{n+1-\sqrt{n+1}}>\frac{1}{n+1}$ for all values of $n$, the series of absolute values is term by term greater than the harmonic series, and thus is divergent. The given series is conditionally convergent.

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01:12

Problem 10

Investigate each of the following series for convergence or divergence:
(a) $\frac{1}{3}+\frac{1}{6}+\frac{1}{11}+\cdots+\frac{1}{2^n+n}+\cdots$
(f) $1+\frac{1}{\sqrt[3]{2}}+\frac{1}{\sqrt[3]{3}}+\cdots+\frac{1}{\sqrt[3]{n}}+\cdots$
(b) $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2 n}+\cdots$
(g) $\frac{1}{2}+\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}+\cdots+\frac{1}{n \cdot 2^n}+\cdots$
(c) $1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2 n-1}+\cdots$
(h) $\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\cdots+\frac{n}{n+1}+\cdots$
(d) $\frac{2}{1 !}+\frac{2^2}{2 !}+\frac{2^3}{3 !}+\cdots+\frac{2^n}{n !}+\cdots$
(i) $\frac{1}{1 \cdot 3}+\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\cdots+\frac{1}{(2 n-1)(2 n+1)}+\cdots$
(e) $2+\frac{1}{2}+\frac{8}{27}+\frac{1}{4}+\cdots+\frac{2^n}{n^3}+\cdots$
(j) $1+\frac{2^2+1}{2^3+1}+\frac{3^2+1}{3^3+1}+\cdots+\frac{n^2+1}{n^3+1}+\cdots$

Linh Vu
Linh Vu
Numerade Educator
01:28

Problem 11

Investigate the following alternating series for convergence or divergence:
(a) $\frac{1}{4}-\frac{1}{10}+\frac{1}{28}-\frac{1}{82}+\cdots$
(f) $\frac{2}{3}-\frac{3}{4} \cdot \frac{1}{2}+\frac{4}{5} \cdot \frac{1}{3}-\frac{5}{6} \cdot \frac{1}{4}+\cdots$
(b) $2-\frac{3}{2}+\frac{4}{3}-\frac{5}{4}+\cdots$
(g) $\frac{2}{2 \cdot 3}-\frac{2^2}{3 \cdot 4}+\frac{2^3}{4 \cdot 5}-\frac{2^4}{5 \cdot 6}+\cdots$
(c) $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
(h) $2-\frac{2^3}{3 !}+\frac{2^5}{5 !}-\frac{2^7}{7 !}+\cdots$
(d) $\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\cdots$
(e) $\frac{1}{4}-\frac{3}{6}+\frac{5}{8}-\frac{7}{10}+\cdots$

Nick Johnson
Nick Johnson
Numerade Educator