9.4E: Exercises for Section 9.4

Use the Comparison Test to determine whether each series in exercises 1 - 13 converges or diverges.

1) $\displaystyle \sum^∞_{n=1}a_n$ where $a_n=\dfrac{2}{n(n+1)}$

2) $\displaystyle \sum^∞_{n=1}a_n$ where $a_n=\dfrac{1}{n(n+1/2)}$

Answer

Converges by comparison with $\dfrac{1}{n^2}$.

3) $\displaystyle \sum^∞_{n=1}\frac{1}{2(n+1)}$

4) $\displaystyle \sum^∞_{n=1}\frac{1}{2n−1}$

Answer

Since $2n−1<2n$, we have $\displaystyle \frac{1}{2n} < \frac{1}{2n-1} \to \sum^∞_{n=1}\frac{1}{2n} < \sum^∞_{n=1}\frac{1}{2n-1}$.
But the series on the left is just half of the harmonic series, since $\displaystyle \sum^∞_{n=1}\frac{1}{2n} = \tfrac{1}{2}\sum^∞_{n=1}\frac{1}{n}$.  This clearly diverges, so the given series also diverges, by the Comparison Test.

5) $\displaystyle \sum^∞_{n=2}\frac{1}{(n\ln n)^2}$

6) $\displaystyle \sum^∞_{n=1}\frac{n!}{(n+2)!}$

Answer

$a_n=1/(n+1)(n+2)<1/n^2.$ Converges by comparison with $p$-series, $p=2>1$.

7) $\displaystyle \sum^∞_{n=1}\frac{1}{n!}$

8) $\displaystyle \sum^∞_{n=1}\frac{\sin(1/n)}{n}$

Answer

$\sin(1/n)≤1/n,$ so converges by comparison with $p$-series, $p=2>1$.

9) $\displaystyle \sum_{n=1}^∞\frac{\sin^2n}{n^2}$

10) $\displaystyle \sum_{n=1}^∞\frac{\sin(1/n)}{\sqrt{n}}$

Answer

$\sin(1/n)≤1,$ so converges by comparison with $p$-series, $p=3/2>1.$

11) $\displaystyle \sum^∞_{n=1}\frac{n^{1.2}−1}{n^{2.3}+1}$

12) $\displaystyle \sum^∞_{n=1}\frac{\sqrt{n+1}−\sqrt{n}}{n}$

Answer

Since $\sqrt{n+1}−\sqrt{n}=1/(\sqrt{n+1}+\sqrt{n})≤2/\sqrt{n},$ series converges by comparison with $p$-series for $p=1.5>1$.

13) $\displaystyle \sum^∞_{n=1}\frac{\sqrt[4]{n}}{\sqrt[3]{n^4+n^2}}$

Use the Limit Comparison Test to determine whether each series in exercises 14 - 28 converges or diverges.

14) $\displaystyle \sum^∞_{n=1}\left(\frac{\ln n}{n}\right)^2$

Answer

Converges by limit comparison with $p$-series for $p>1$.

15) $\displaystyle \sum^∞_{n=1}\left(\frac{\ln n}{n^{0.6}}\right)^2$

16) $\displaystyle \sum^∞_{n=1}\frac{\ln\left(1+\frac{1}{n}\right)}{n}$

Answer

Converges by limit comparison with $p$-series, $p=2>1.$

17) $\displaystyle \sum^∞_{n=1}\ln\left(1+\frac{1}{n^2}\right)$

18) $\displaystyle \sum^∞_{n=1}\frac{1}{4^n−3^n}$

Answer

Converges by limit comparison with $4^{−n}$.

19) $\displaystyle \sum^∞_{n=1}\frac{1}{n^2−n\sin n}$

20) $\displaystyle \sum^∞_{n=1}\frac{1}{e^{(1.1)n}−3^n}$

Answer

Converges by limit comparison with $1/e^{1.1n}$.

21) $\displaystyle \sum^∞_{n=1}\frac{1}{e^{(1.01)n}−3^n}$

22) $\displaystyle \sum^∞_{n=1}\frac{1}{n^{1+1/n}}$

Answer

Diverges by limit comparison with harmonic series.

23) $\displaystyle \sum^∞_{n=1}\frac{1}{2^{1+1/n}}{n^{1+1/n}}$

24) $\displaystyle \sum^∞_{n=1}\left(\frac{1}{n}−\sin\left(\frac{1}{n}\right)\right)$

Answer

Converges by limit comparison with $p$-series, $p=3>1$.

25) $\displaystyle \sum^∞_{n=1}\left(1−\cos\left(\frac{1}{n}\right)\right)$

26) $\displaystyle \sum^∞_{n=1}\frac{1}{n}\left(\tan^{−1}n−\frac{π}{2}\right)$

Answer

Converges by limit comparison with $p$-series, $p=3>1$.

27) $\displaystyle \sum^∞_{n=1}\left(1−\frac{1}{n}\right)^{n.n}$ (Hint: $\left(1−\dfrac{1}{n}\right)^n→1/e.$)

28) $\displaystyle \sum^∞_{n=1}\left(1−e^{−1/n}\right)$ (Hint: $1/e≈(1−1/n)^n,$ so $1−e^{−1/n}≈1/n.$)

Answer

Diverges by limit comparison with $1/n$.

29) Does $\displaystyle \sum^∞_{n=2}\frac{1}{(\ln n)^p}$ converge if $p$ is large enough? If so, for which $p?$

30) Does $\displaystyle \sum^∞_{n=1}\left(\frac{\ln n}{n}\right)^p$ converge if $p$ is large enough? If so, for which $p?$

Answer

Converges for $p>1$ by comparison with a $p$ series for slightly smaller $p$.

31) For which $p$ does the series $\displaystyle \sum^∞_{n=1}2^{pn}/3^n$ converge?

32) For which $p>0$ does the series $\displaystyle \sum^∞_{n=1}\frac{n^p}{2^n}$ converge?

Answer

Converges for all $p>0$.

33) For which $r>0$ does the series $\displaystyle \sum^∞_{n=1}\frac{r^{n^2}}{2^n}$ converge?

34) For which $r>0$ does the series $\displaystyle \sum^∞_{n=1}\frac{2^n}{r^{n^2}}$ converge?

Answer

Converges for all $r>1$. If $r>1$ then $r^n>4$, say, once $n>\ln(2)/\ln(r)$ and then the series converges by limit comparison with a geometric series with ratio $1/2$.

35) Find all values of $p$ and $q$ such that $\displaystyle \sum^∞_{n=1}\frac{n^p}{(n!)^q}$ converges.

36) Does $\displaystyle \sum^∞_{n=1}\frac{\sin^2(nr/2)}{n}$ converge or diverge? Explain.

Answer

The numerator is equal to $1$ when $n$ is odd and $0$ when $n$ is even, so the series can be rewritten $\displaystyle \sum^∞_{n=1}\frac{1}{2n+1},$ which diverges by limit comparison with the harmonic series.

37) Explain why, for each $n$, at least one of ${|\sin n|,|\sin(n+1)|,...,|\sin(n+6)|}$ is larger than $1/2$. Use this relation to test convergence of $\displaystyle \sum^∞_{n=1}\frac{|\sin n|}{\sqrt{n}}$.

38) Suppose that $a_n≥0$ and $b_n≥0$ and that $\displaystyle \sum_{n=1}^∞a^2_n$ and $\displaystyle \sum_{n=1}^∞b^2_n$ converge. Prove that $\displaystyle \sum_{n=1}^∞a_nb_n$ converges and $\displaystyle \sum_{n=1}^∞a_nb_n≤\frac{1}{2}\left(\sum_{n=1}^∞a^2_n+\sum_{n=1}^∞b^2_n\right)$.

Answer

$(a−b)^2=a^2−2ab+b^2$ or $a^2+b^2≥2ab$, so convergence follows from comparison of $2a_nb_n$ with $a^2_n+b^2_n.$ Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.

39) Does $\displaystyle \sum_{n=1}^∞2^{−\ln\ln n}$ converge? (Hint: Write $2^{\ln\ln n}$ as a power of $\ln n$.)

40) Does $\displaystyle \sum_{n=1}^∞(\ln n)^{−\ln n}$ converge? (Hint: Use $t=e^{\ln(t)}$ to compare to a $p$−series.)

Answer

$(\ln n)^{−\ln n}=e^{−\ln(n)\ln\ln(n)}.$ If $n$ is sufficiently large, then $\ln\ln n>2,$ so $(\ln n)^{−\ln n}<1/n^2$, and the series converges by comparison to a $p$−series.

41) Does $\displaystyle \sum_{n=2}^∞(\ln n)^{−\ln\ln n}$ converge? (Hint: Compare $a_n$ to $1/n$.)

42) Show that if $a_n≥0$ and $\displaystyle \sum_{n=1}^∞a_n$ converges, then $\displaystyle \sum_{n=1}^∞a^2_n$ converges. If $\displaystyle \sum_{n=1}^∞a^2_n$ converges, does $\displaystyle \sum_{n=1}^∞a_n$ necessarily converge?

Answer

$a_n→0,$ so $a^2_n≤|a_n|$ for large $n$. Convergence follows from limit comparison. $\displaystyle \sum_{n=1}^∞\frac{1}{n^2}$ converges, but $\displaystyle \sum_{n=1}^∞\frac{1}{n}$ does not, so the fact that $\displaystyle \sum_{n=1}^∞a^2_n$ converges does not imply that $\displaystyle \sum_{n=1}^∞a_n$ converges.

43) Suppose that $a_n>0$ for all $n$ and that $\displaystyle \sum_{n=1}^∞a_n$ converges. Suppose that $b_n$ is an arbitrary sequence of zeros and ones. Does $\displaystyle \sum_{n=1}^∞a_nb_n$ necessarily converge?

44) Suppose that $a_n>0$ for all $n$ and that $\displaystyle \sum_{n=1}^∞a_n$ diverges. Suppose that $b_n$ is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does $\displaystyle \sum_{n=1}^∞a_nb_n$ necessarily diverge?

Answer

No. $\displaystyle \sum_{n=1}^∞1/n$ diverges. Let $b_k=0$ unless $k=n^2$ for some $n$. Then $\displaystyle \sum_kb_k/k=\sum1/k^2$ converges.

45) Complete the details of the following argument: If $\displaystyle \sum_{n=1}^∞\frac{1}{n}$ converges to a finite sum $s$, then $\dfrac{1}{2}s=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+⋯$ and $s−\dfrac{1}{2}s=1+\dfrac{1}{3}+\dfrac{1}{5}+⋯.$ Why does this lead to a contradiction?

46) Show that if $a_n≥0$ and $\displaystyle \sum_{n=1}^∞a^2_n$ converges, then $\displaystyle \sum_{n=1}^∞\sin^2(a_n)$ converges.

Answer

$|\sin t|≤|t|,$ so the result follows from the comparison test.

47) Suppose that $a_n/b_n→0$ in the comparison test, where $a_n≥0$ and $b_n≥0$. Prove that if $\displaystyle \sum b_n$ converges, then $\displaystyle \sum a_n$ converges.

48) Let $b_n$ be an infinite sequence of zeros and ones. What is the largest possible value of $\displaystyle x=\sum_{n=1}^∞b_n/2^n$?

Answer

By the comparison test, $\displaystyle x=\sum_{n=1}^∞b_n/2^n≤\sum_{n=1}^∞1/2^n=1.$

49) Let $d_n$ be an infinite sequence of digits, meaning $d_n$ takes values in $\{0,1,…,9\}$. What is the largest possible value of $\displaystyle x=\sum_{n=1}^∞d_n/10^n$ that converges?

50) Explain why, if $x>1/2,$ then $x$ cannot be written $\displaystyle x=\sum_{n=2}^∞\frac{b_n}{2^n}\quad (b_n=0\;\text{or}\;1,\;b_1=0).$

Answer

If $b_1=0,$ then, by comparison, $\displaystyle x≤\sum_{n=2}^∞1/2^n=1/2.$

51) [T] Evelyn has a perfect balancing scale, an unlimited number of $1$-kg weights, and one each of $1/2$-kg, $1/4$-kg, $1/8$-kg, and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?

52) [T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of $1$-kg weights, and nine each of $0.1$-kg, $0.01$-kg, $0.001$-kg, and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?

Answer

Yes. Keep adding $1$-kg weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the $1$-kg weights, and add $0.1$-kg weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last $0.1$-kg weight. Start adding $0.01$-kg weights. If it balances, stop. If it tips to the side with the weights, remove the last $0.01$-kg weight that was added. Continue in this way for the $0.001$-kg weights, and so on. After a finite number of steps, one has a finite series of the form $\displaystyle A+\sum_{n=1}^Ns_n/10^n$ where $A$ is the number of full kg weights and $d_n$ is the number of $1/10^n$-kg weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the $N^{\text{th}}$ partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most $1/10^N$.

53) The series $\displaystyle \sum_{n=1}^∞\frac{1}{2n}$ is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which $n$ is odd. Let $m>1$ be fixed. Show, more generally, that deleting all terms $1/n$ where $n=mk$ for some integer $k$ also results in a divergent series.

54) In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from $\displaystyle \sum_{n=1}^∞\frac{1}{n}$ by removing any term $1/n$ if a given digit, say $9$, appears in the decimal expansion of $n$. Argue that this depleted harmonic series converges by answering the following questions.

a. How many whole numbers $n$ have $d$ digits?

b. How many $d$-digit whole numbers $h(d)$. do not contain $9$ as one or more of their digits?

c. What is the smallest $d$-digit number $m(d)$?

d. Explain why the deleted harmonic series is bounded by $\displaystyle \sum_{d=1}^∞\frac{h(d)}{m(d)}$.

e. Show that $\displaystyle \sum_{d=1}^∞\frac{h(d)}{m(d)}$ converges.

Answer

a. $10^d−10^{d−1}<10^d$
b. $h(d)<9^d$
c. $m(d)=10^{d−1}+1$
d. Group the terms in the deleted harmonic series together by number of digits. $h(d)$ bounds the number of terms, and each term is at most $\frac{1}{m(d)}.$
Then $\displaystyle \sum_{d=1}^∞h(d)/m(d)≤\sum_{d=1}^∞9^d/(10)^{d−1}≤90$. One can actually use comparison to estimate the value to smaller than $80$. The actual value is smaller than $23$.

55) Suppose that a sequence of numbers $a_n>0$ has the property that $a_1=1$ and $a_{n+1}=\dfrac{1}{n+1}S_n$, where $S_n=a_1+⋯+a_n$. Can you determine whether $\displaystyle \sum_{n=1}^∞a_n$ converges? (Hint: $S_n$ is monotone.)

56) Suppose that a sequence of numbers $a_n>0$ has the property that $a_1=1$ and $a_{n+1}=\dfrac{1}{(n+1)^2}S_n$, where $S_n=a_1+⋯+a_n$. Can you determine whether $\displaystyle \sum_{n=1}^∞a_n$ converges? (Hint: $S_2=a_2+a_1=a_2+S_1=a_2+1=1+1/4=(1+1/4)S_1, S_3=\dfrac{1}{3^2}S_2+S_2=(1+1/9)S_2=(1+1/9)(1+1/4)S_1$, etc. Look at $\ln(S_n)$, and use $\ln(1+t)≤t, t>0.$)

Answer

Continuing the hint gives $S_N=(1+1/N^2)(1+1/(N−1)^2…(1+1/4)).$ Then $\ln(S_N)=\ln(1+1/N^2)+\ln(1+1/(N−1)^2)+⋯+\ln(1+1/4).$ Since $\ln(1+t)$ is bounded by a constant times $t$, when $0<t<1$ one has $\displaystyle \ln(S_N)≤C\sum_{n=1}^N\frac{1}{n^2}$, which converges by comparison to the $p$-series for $p=2$.