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 \(p=2\).