4.4E: Excersieses

Exercise \(\PageIndex{1}\)

Use the comparison test to determine whether the following series converge.

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

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

Answer

Solution: Converges by comparison with \(\displaystyle 1/n^2\).

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

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

Answer

Solution: Diverges by comparison with harmonic series, since \(\displaystyle 2n−1≥n.\)

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

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

Answer

Solution: \(\displaystyle a_n=1/(n+1)(n+2)<1/n^2.\) Converges by comparison with p-series, \(\displaystyle p=2\).

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

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

Answer

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

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

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

Answer

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

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

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

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

Exercise \(\PageIndex{2}\)

Use the limit comparison test to determine whether each of the following series converges or diverges.

1) \(\displaystyle \sum^∞_{n=1}(\frac{lnn}{n})^2\)

Answer

Solution: Converges by limit comparison with p-series for \(\displaystyle p>1\).

2) \(\displaystyle \sum^∞_{n=1}(\frac{lnn}{n^{0.6}})^2\)

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

Answer

Solution: Converges by limit comparison with p-series, \(\displaystyle p=2.\

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

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

Answer

Solution: Converges by limit comparison with \(\displaystyle 4^{−n}\).

6) \(\displaystyle \sum^∞_{n=1}\frac{1}{n^2−nsinn}\)

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

Answer

Solution: Converges by limit comparison with \(\displaystyle 1/e^{1.1n}\).

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

9) \(\displaystyle \sum^∞_{n=1}\frac{1}{n^{1+1/n}}\)

Answer

Solution: Diverges by limit comparison with harmonic series.

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

11) \(\displaystyle \sum^∞_{n=1}(\frac{1}{n}−sin(\frac{1}{n}))\)

Answer

Solution: Converges by limit comparison with p-series, \(\displaystyle p=3\).

12) \(\displaystyle \sum^∞_{n=1}(1−cos(\frac{1}{n}))\)

13) \(\displaystyle \sum^∞_{n=1}\frac{1}{n}(tan^{−1}n−\frac{π}{2})\)

Answer

Solution: Converges by limit comparison with p-series, \(\displaystyle p=3\).

15) \(\displaystyle \sum^∞_{n=1}(1−\frac{1}{n})^{n.n}\) (Hint:\(\displaystyle (1−\frac{1}{n})^n→1/e.\))

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

Answer

Solution: Diverges by limit comparison with \(\displaystyle 1/n\).

Exercise \(\PageIndex{3}\)

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

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

Answer

Solution: Converges for \(\displaystyle p>1\) by comparison with a \(\displaystyle p\) series for slightly smaller \(\displaystyle p\).

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

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

Answer

Solution: Converges for all \(\displaystyle p>0\).

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

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

Answer

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

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

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

Answer

Solution: The numerator is equal to \(\displaystyle 1\) when \(\displaystyle n\) is odd and \(\displaystyle 0\) when \(\displaystyle n\) 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.

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

10) Suppose that \(\displaystyle a_n≥0\) and \(\displaystyle b_n≥0\) 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 \(\displaystyle \sum_{n=1}^∞a_nb_n≤\frac{1}{2}(\sum_{n=1}^∞a^2_n+\sum_{n=1}^∞b^2_n)\).

Answer

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

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

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

Answer

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

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

Exercise \(\PageIndex{4}\)

1) Show that if \(\displaystyle a_n≥0\) 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

Solution: \(\displaystyle a_n→0,\) so \(\displaystyle a^2_n≤|a_n|\) for large \(\displaystyle n\). Convergence follows from limit comparison. \(\displaystyle \sum1/n^2\) converges, but \(\displaystyle \sum1/n\) 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.

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

3) Suppose that \(\displaystyle a_n>0\) for all \(\displaystyle n\) and that \(\displaystyle \sum_{n=1}^∞a_n\) diverges. Suppose that \(\displaystyle b_n\) 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

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

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

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

Answer

Solution: \(\displaystyle |sint|≤|t|,\) so the result follows from the comparison test.

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

7) Let \(\displaystyle b_n\) 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

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

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

9) Explain why, if \(\displaystyle x>1/2,\) then \(\displaystyle x\) cannot be written \(\displaystyle x=\sum_{n=2}^∞\frac{b_n}{2^n}(b_n=0or1,b_1=0).\)

Answer

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

Exercise \(\PageIndex{5}\)

1) [T] Evelyn has a perfect balancing scale, an unlimited number of \(\displaystyle 1-kg\) weights, and one each of \(\displaystyle 1/2-kg,1/4-kg,1/8-kg,\) 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?

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

Answer

Solution: Yes. Keep adding \(\displaystyle 1-kg\) 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 \(\displaystyle 1-kg\) weights, and add \(\displaystyle 0.1-kg\) weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last \(\displaystyle 0.1-kg\) weight. Start adding \(\displaystyle 0.01-kg\) weights. If it balances, stop. If it tips to the side with the weights, remove the last \(\displaystyle 0.01-kg\) weight that was added. Continue in this way for the \(\displaystyle 0.001-kg\) 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 \(\displaystyle A\) is the number of full kg weights and \(\displaystyle d_n\) is the number of \(\displaystyle 1/10^n-kg\) weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the Nth partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most \(\displaystyle 1/10^N\).

3) 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 \(\displaystyle n\) is odd. Let \(\displaystyle m>1\) be fixed. Show, more generally, that deleting all terms \(\displaystyle 1/n\) where \(\displaystyle n=mk\) for some integer \(\displaystyle k\) also results in a divergent series.

4) 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 \(\displaystyle 1/n\) if a given digit, say \(\displaystyle 9\), appears in the decimal expansion of \(\displaystyle n\).Argue that this depleted harmonic series converges by answering the following questions.

a. How many whole numbers \(\displaystyle n\) have \(\displaystyle d\) digits?

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

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

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

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

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

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

Answer

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