Chapter 3 Numerical Sequences and Series
Exercise 1
(By ghostofgarborg) Assume . Choose , and let be such that when . Then whenever . This implies convergence of .
The converse is not true. Let . This sequence does not converge, even though does.
Exercise 2
(By ghostofgarborg) Note that , and therefore
where . By theorem 3.19, .
(By Dan kyp44 Whitman) We can calculate this limit directly:
noting that clearly .
Exercise 3
(By ghostofgarborg) Note that if , , so the sequence is bounded. Also note that if , then , so the sequence is monotonically increasing. It therefore converges by thm 3.14.
Exercise 4
(By ghostofgarborg) We observe that if and , then and fulfills both the initial condition and the recursion, so that this is a closed form of the sequence. Note that , , and that any subsequence of contains either a subsequence of or a subsequence of . Consequently, any convergent subsequence converges to either or . We get
Exercise 5
(By ghostofgarborg) If and , the result is trivial. If and , then is bounded above by and given a number the latter of which is finite. Therefore .
We can therefore assume that both and are finite. There is a subsequence such that . By choosing this subsequence right, i.e. if necessary passing to another subsequence, we can make sure that is convergent. Then is convergent, so
Exercise 6
(By ghostofgarborg)
(a) Since the series telescopes, . This shows that it diverges to .
(b) We haveConverges by the comparison test (thm. 3.25).
(c) We have Converges by the root test.
(d) When , violating the necessary condition that . In this case,
is divergent. When , and converges by comparison with the geometric series .
Exercise 7
(By ghostofgarborg) Assume . We know that converges. Let be the number it converges to. Note that is monotonic. By the Cauchy-Schwarz inequality This implies that is bounded, hence converges.
Exercise 8
(By ghostofgarborg) We first note that thm 3.42 holds for a monotonously increasing sequence whose limit is 0 as well, since then fulfills the criteria of the theorem, and .
If converges, the partial sums form a bounded sequence. If is monotonic and bounded it converges to a number , and we get that The first sum on the right hand side converges by thm 3.42 and the observation above. The second sum converges because does. Consequently the left hand side converges.
Exercise 9
(By ghostofgarborg)
(a) We haveSo .
(b) We have So .
(c) We have So .
(d) We have So .
Exercise 10
(By ghostofgarborg) If and for an infinite number of values of , then , and does not tend to . This makes the series divergent when .
Exercise 11
(By ghostofgarborg)
(a) If , then . If , then . Therefore If is infinite, this clearly diverges. Otherwise, there is an such that implies , in which case the series diverges by comparison to .
(b) Since is monotonically increasing, whenever Consequently, Since is increasing and unbounded, it is possible to choose such that is arbitrarily close to . This shows that is not Cauchy, and therefore not convergent.
(c) Since is monotonically increasing, we have that , so we can deduce that Since is bounded so is monotonic and bounded, hence convergent.
(d) Since converges. The series , however, might converge or diverge. Let , and it is clear that it diverges. Let whenever is a square and otherwise. This series clearly diverges, since the terms do not tend to 0 as . Then and the series therefore converges.
Exercise 12
(By ghostofgarborg)
(a) Since is monotonically decreasing Since , given any we can find an such that is arbitrarily close to . This implies that is not Cauchy, hence not convergent.
(b) We have Since the series converges by the comparison test.
Exercise 13
(By ghostofgarborg) Since and converge, so does their Cauchy product . Let be the Cauchy product of and . Then , and as it is majorized by a convergent series, converges . I.e. converges absolutely.
Exercise 14
(By ghostofgarborg)
(a) Choose and such that for all . Then choose such that Then whenever This shows that converges to , i.e. that .
(b) Let . Then does not converge, but converges to .
(c) Yes. Let , , and when is not a power of .
Let . Then , and the series is unbounded, so . Howeverwhere the right hand side tends to 0. This shows that .
(d) Note thatso that
Note that the right hand side is the th arithmetic mean of the sequence , which by assumption and our result in (a) converges. Therefore is convergent, and since is convergent by assumption, is convergent.
(e) We have which is the desired equality. By assumption , so that The rest of the argument is covered in sufficient detail in the text.
Baby Rudin 数学分析原理完整第三章习题解答