3.7:4. A necessary and sufficient condition for convergence to be testable by sampling. (d : 1, 4. > 3.7:3) The preceding exercise found a condition on an increasing sequence of positive integers nk which is sufficient for the convergence of any decreasing series Σ an of positive terms to be determinable from the
‘‘sampling’’ of values an
k. However, part (a) below shows that this condition is not necessary. Part (b) will establish a condition that is both necessary and sufficient.
(a) Let (nk) be the increasing sequence of positive integers such that {nk} = { 2i ! i > 0} ∪ { 2i+1 ! i > 0}. (Explicitly, n2k –1 = 2k, n2k = 2k+1.) Show that if (an) and (bn) are decreasing sequence of positive terms such that bn
k = an
k for all k, then Σ∞n = 1 bn converges if and only if Σ∞n = 1 an does.
(Hint: The above sequence of integers has a subsequence to which the preceding exercise is applicable.) The above example suggests that we might incorporate the existence of appropriate subsequences into our criterion. And in fact, doing so gives the desired necessary and sufficient condition:
(b) Show that for positive integers 1 ≤ n1 < n2 < n3 < ... , the following three conditions are equivalent:
(i) For every decreasing sequence of positive terms (an), the series Σ an converges if and only if the series Σ(nk+1– nk) an
k converges.
...
Answer to True/False question 3.7:0. (a)F.
(ii) If (an) and (bn) are decreasing sequence of positive terms such that bn
k = an
k for all k, then the series Σ an converges if and only if Σ bn does.
(iii) The set of ratios nk+1⁄ nk is bounded.
(iv) (nk) has a subsequence (nm
k) such that the set of ratios (nm
k+1 – nm
k) ⁄ (nm
k – nm
k –1) is bounded.
Suggestion: Prove (i)⇒(ii)⇒(iii)⇒(iv)⇒(i). The second implication is the hardest; I suggest proving it in contrapositive form, using the idea of 3.7:2. For the third implication, to get the subsequence (nm
k), suppose mk has been chosen; then let mk+1 be the least integer such that nm
k+1 ≥ 2 nm
k. 3.7:5. Series that can be tested for convergence by looking at the terms a
22k. (d : 2. > 3.7:2)
The fallacious argument discussed in 3.7:2(b) can in fact be made to work, if we just add an additional condition to our series.
Namely, show that if (an) is a sequence of real numbers satisfying the stronger condition a1 ≥ 2 a2 ≥ ... ≥ n an ≥ ... , then Σ an converges if and only if Σ 2k22ka
22k does. Deduce that if (bn) is another sequence with the same property, and for all k, a
22k = b
22k, then Σ an converges if and only if Σ bn does.
3.7:6. The series of distances associated with a convergent sequence. (d : 2)
Suppose ( pn) is a sequence in a metric space X which converges to a point p.
(a) Show that d ( p1, p) ≤ Σ∞
n = 1 d ( pn, pn+1).
(b) Show that for every ε> 0 there exists a subsequence ( pn
k) with n1 = 1 such that Σk d ( pn
k, pn
k+1) < d ( p1, p) +ε.
Now suppose ( pn) is an arbitrary sequence of points in X.
(c) Show that if the series Σ∞
n = 1 d ( pn, pn+1) converges, then ( pn) is a Cauchy sequence; but that the converse is not true. (Hint for the last part: Look for a counterexample in a metric space we are very familiar with.)
3.8. THE NUMBER e. (pp.63-65)
Relevant exercises in Rudin: None Exercises not in Rudin: None 3.9. THE ROOT AND RATIO TESTS. (pp.65-69)
Relevant exercises in Rudin: None Exercises not in Rudin:
3.9:0. Say whether each of the following statements is true or false.
(a) If (an) is a sequence such that lim supn→∞|an+1 ⁄ an| = 1⁄2, then Σ∞
n = 1 an converges.
(b) If (an) is a sequence such that lim supn→∞|an+1 ⁄ an| = 2, then Σ∞
n = 1 an diverges.
3.9:1. Sharper convergence tests analogous to the ratio and root tests. (d : 1, 3, 2, 2, 4, 2)
The idea behind the ratio and root tests is to compare a given series with series of the form Σ c xn, whose behavior we know for each value of x. But we have seen that series of the form Σ n– p are too delicate for those tests to work on. Is it possible to devise tests that would similarly compare a series with series of the form Σ n– p?
Yes; such tests are given below. The idea, analogous to the idea behind the ratio and root tests, is to find a limit-formula that, given a sequence of the form Σn– p, will determine the value of p, and then apply it to more general series.
The analog of the root test, based on considering the ‘‘long-term’’ change in magnitude of the terms of
the series, is given in part (b), and is relatively easy to state and prove if we again allow ourselves to assume basic properties of the logarithm function. The analog of the ratio test, based on considering the relation between successive terms, is noted in part (e). Like the ratio test, it is generally easier to use than the other criterion, but cannot handle series whose terms don’t decrease regularly. It is, unfortunately, hard to prove without the use of differentiation, which we have not yet defined. (The one way I see that one could prove it at this stage is by taking a rational number j ⁄ k between p and 1, and using inequalities obtained from the binomial theorem for exponents j and k.) However, I will state both tests here for their interest.
(a) If p is any positive real number, and we let an = n– p, show that – limn→∞ (log an) ⁄ (log n) = p.
(b) Let (an) be any sequence of positive terms, and let p = – lim sup (log an) ⁄ (log n). Show that if p > 1, then Σan converges.
(c) What happens when we apply this test to a series which the root test shows to converge?
(d) For p again any positive real number and an = n– p, show that limn→∞ n (1 – (an+1 ⁄ an)) = p.
(e) Let (an) be any sequence of positive terms, and let p = lim inf n (1 – (an+1 ⁄ an)). Show that if p > 1, then Σan converges.
(f ) What happens when we apply this test to a series which the ratio test shows to converge?
3.9:2. A slightly modified ratio test. (d : 3,1)
(a) Show that a series Σ an converges if for some positive integer c one has lim supn→∞ |an+c ⁄ an| < 1.
(b) Show that this condition is satisfied for c = 2 by both the series of Examples 3.35 (p.67).
3.10. POWER SERIES. (pp.69-70)
Relevant exercises in Rudin:
3:R9. Finding the radii of convergence of some power series. (d : 2)
3:R10. The radius of convergence of a power series with integer coefficients. (d : 2) Exercises not in Rudin:
3.10:0. Say whether the following statement is true or false.
(a) If a power series Σ∞
n = 1 cnzn converges at z = 1 + 3 i, then it also converges at z = 2 + 2 i.
3.10:1. Power series and the ratio test. (d : 1)
Let Σcnzn be a power series in which all the coefficients cn are nonzero.
(a) Show that if the sequence ( |cn⁄ cn+1| ) approaches either a real number or +
∞
, then that limit is equal to the radius of convergence of the power series.(b) If we do not assume that ( |cn⁄ cn+1| ) approaches a limit (real or infinite), obtain an inequality relating lim sup |cn⁄ cn+1| and the radius of convergence of the given power series.
(c) Likewise, obtain an inequality relating lim inf |cn⁄ cn+1| and the radius of convergence of the power series.
(Hint: You do not have to give any nontrivial arguments to get parts (a)-(c); everything can be obtained easily from results in Rudin.)
(d) Consider the series Σcnzn where for each nonnegative integer k, we let c2 k = c2 k+1 = 2k. Determine the radius of convergence of this series using Theorem 3.39, determine the upper and lower limits described in (b) and (c) above, and verify for this series the inequalities proved there.
...
Answers to True/False question 3.9:0. (a)T. (b)F.
3.11. SUMMATION BY PARTS. (pp.70-71)
Relevant exercises in Rudin:
3:R7. If Σ an converges, so does Σ0a--- ⁄ n. (d: 3)n
Suggestion: Write an = bn2 and use the Schwarz inequality. Note that the page ends with a very short line which gives one of the hypotheses of the exercise.
3:R8. Tweaking the hypotheses of Theorem 3.42. (d : 2)
Exercises not in Rudin:
3.11:0. Say whether each of the following statements is true or false.
(a) The series Σ (–1)n(1 + n–1) is convergent.
(b) The series 1 ⁄ 1 + 1 ⁄ 2 – 2 ⁄ 3 + 1 ⁄ 4 + 1 ⁄ 5 – 2 ⁄ 6 + ... is convergent, where the nth denominator is n, and the nth numerator is 1 if n is not divisible by 3 and – 2 if n is divisible by 3.
3.11:1. Symmetrizing the ‘‘summation by parts’’ formula. (d : 2)
In display (20) on p.70 of Rudin, the two sums have different ranges of summation. Obtain a similar formula in which the two sums are over the same range of values of n, by adding to the sum on the right-hand side the missing n = q term, subtracting the same term from the end of the formula, and simplifying the result by canceling a pair of terms. (The formula you get should, like (20), have a summation on each side of the equality, and a pair of lone terms.)
3.11:2. A generalization of Σ ( –1)nn–1. (d : 2)
Let a1, a2, ... , ad be a finite sequence of complex numbers, and extend it to an infinite sequence by making it periodic of period d, i.e., letting ad k + i = ai for all positive integers k and all i∈{1, ... , d } . Prove that Σ∞
n = 1 an⁄ n converges if and only if a1+ a2+ ... + ad = 0.
3.11:3. A version of Theorem 3.42 with complex bn. (d : 2)
In Theorem 3.42, p.70, the an can be complex numbers, but the bn are necessarily real, by Rudin’s convention that inequality signs such as ‘‘≥’’ are only written between real numbers. However, we can replace condition (b) of that theorem with a condition applicable to complex numbers as well:
(a) Show that Theorem 3.42 remains true if condition (b) of that theorem is replaced by the assumption that Σ| bn+1 – bn| converges.
(b) Show that the assumption of part (a) above does in fact hold whenever conditions (b) and (c) of Theorem 3.42 hold. (Thus, the result proved in part (a) includes that theorem.)
Remark: From the above generalization of Theorem 3.42, we can immediately get the corresponding version of Theorem 3.44. The same method of proof that gives (a) above also easily gives a version where (an) and (bn) are replaced by sequences in Rk, ( an) and ( bn), and the conclusion is that Σ an· bn converges.
3.12. ABSOLUTE CONVERGENCE. (pp.71-72)
Relevant exercises in Rudin: None Exercises not in Rudin:
3.12:0. Say whether the following statement is true or false.
(a) The series 1 ⁄ 1 + 1 ⁄ 2 – 2 ⁄ 3 + 1 ⁄ 4 + 1 ⁄ 5 – 2 ⁄ 6 + ... (cf. 3.11:0(b)) is absolutely convergent.
3.12:1. Achieving absolute convergence by grouping. (d : 1, 2)
(a) Suppose Σn an is a convergent series. Show that for any sequence of integers n1 < n2 < ...
< nk < ... with n1 = 1, if for each positive integer k we define Ak = an
k + an
k+1 + an
k+2 + ... + an
k+1–2 + an
k+1–1,
...
Answer to True/False question 3.10:0. (a)T.
then the series Σk Ak converges, and Σk Ak = Σn an.
(b) Show that for any convergent series Σn an, there exists a sequence n1 < n2 < ... as in (a) such that the series Σk Ak is absolutely convergent. (Hint: For any convergent series Σ cn of positive terms, choose to nk to make Σ Ak converge by the comparison test with Σ ck.)
3.12:2. Power series converge absolutely inside their radii of convergence. (d : 1, 2) Let Σ anzn be a power series with radius of convergence R > 0
(a) Show that for all complex numbers z with | z | < R, the given series converges absolutely.
(b) Show by three examples that for a complex number z with | z | = R, the series Σ anzn may diverge, may converge nonabsolutely, or may converge absolutely.
(In fact, you can find a power series that shows two of the above phenomena at different values of z with | z | = R; but an example of the remaining phenomenon requires a different power series. Do you see why?)
3.13. ADDITION AND MULTIPLICATION OF SERIES. (pp.72-75) Relevant exercise in Rudin:
3:R13. Cauchy products of absolutely convergent series. (d : 2) Exercises not in Rudin:
3.13:0. Say whether the following statement is true or false.
(a) If Σan = A and Σbn = B, and these series converge absolutely, then Σanbn = A B.
3.13:1. Radii of convergence of sum and product series. (d : 2. > 3.12:2)
Suppose Σ an and Σ bn are series, and let their sum in the sense of Theorem 3.47 and their product in the sense of Definition 3.48 be denoted Σ sn and Σ pn respectively.
(a) Show that if r is a real number such that Σ anzn and Σ bnzn both have radius of convergence
≥ r, then Σ snzn and Σ pnzn also have radii of convergence ≥ r.
(Suggestion: Don’t use the ‘‘lim sup’’ formula for the radius of convergence, but its characterization in terms of where power series converge and where they diverge, together with the result of 3.12:2(a).)
(b) Deduce from (a) that if Σ anzn and Σ bnzn have different radii of convergence, then the radius of convergence of Σ snzn is equal to the smaller of those two radii.
(c) Also deduce from (a) that if the radius of convergence of Σ pnzn is less than that of Σ anzn, then the radius of convergence of Σ bnzn is ≤ that of Σ pnzn.
3.14. REARRANGEMENTS. (pp.75-78)
Relevant exercises in Rudin: None Exercises not in Rudin:
3.14:0. Say whether each of the following statements is true or false.
(a) There is a rearrangement of the series Σ(–1)nn–1 which converges to – 2011.
(b) If a series Σan has the property that all of its rearrangements converge, it is absolutely convergent.
(c) If a series Σan has the property that some rearrangement converges, then it itself converges.
3.14:1. Two ways of showing Rudin’s rearrangement of Σ( –1)n +1⁄ n converges. (d : 3) (a) Show that if (an) is a sequence such that limn→∞ an = 0, then Σ∞
n = 0 an converges if and only if Σ∞
k = 0 (a2k + a2k + 1) converges, and that these infinite sums are then equal. In other words, one can test such a series for convergence, and find its sum if this exists, by doing the same for the series gotten by collecting these pairs of successive terms.
...
Answers to True/False question 3.11:0. (a)F. (b)T. Answer to True/False question 3.12:0. (a)F.
(You should give a careful ‘‘ε’’-proof, unless you prove it using some previous result in Rudin.) (b) Obtain the analogous result relating Σ∞
n = 0 an and Σ∞
k = 0 (a3k + a3k + 1 + a3k + 2). Rather than repeating the whole proof, just indicate carefully what changes need to be made in the proof of (a).
(c) Use (b) above to show that the series (23) on p.76 of Rudin converges. (First describe that series precisely.)
(d) Obtain a different proof of the convergence of Rudin’s series (23) by applying Theorem 3.42 with (an) the sequence ‘‘1, 1, – 2, 1, 1, – 2, 1, 1, – 2, ...’’. (You should figure out what (bn) is to be.)
(e) Show that the result of (a) becomes false if the condition limn→∞ an = 0 is removed.
3.14:2. Which series have convergent rearrangements? (d : 3)
Find a simple criterion for a series Σ an of real numbers to have the property there exists a rearrangement Σ ak
n which converges.
(The answer to this question is fairly easy; the answer to the corresponding question for series of complex numbers, equivalently, of points of R2, is much more difficult to state and prove.)
3.14:3. Which rearrangements of terms don’t affect convergence of any series? (d : 1, 3, 4, 5)
We saw in Theorem 3.54 that rearranging the terms of a non-absolutely convergent series can change its behavior drastically. But not all rearrangements can have such effects. Parts (a)-(c) below show that the effects of certain rearrangements are limited in one way or another. Part (d), which is much harder, gives a general criterion for when this happens.
(a) If (an) is a sequence, consider the rearrangement gotten by interchanging successive pairs a2m and a2m + 1. Clearly, this can be written (ak
n), where (kn) is the sequence of integers 2, 1, 4, 3, 6, 5, ... , defined by k2m –1 = 2m, k2m = 2m –1 (m = 1, 2, ... ).
Show that if Σ an or Σ ak
n is a convergent series, then so is the other, and Σ ak
n = Σ an.
(b) If (an) is a sequence, consider the rearrangement gotten by breaking it into blocks whose lengths are successive powers of 2, i.e., (a
2i, ... , a
2i +1–1), and within each such block, collecting all the even-subscripted terms before the odd-subscripted ones, but otherwise preserving their order. This rearrangement can be written (ak
n), for an appropriate sequence (kn) of integers, whose first 16 terms are as follows. (I use extra space to make visible the separation into ‘‘blocks’’.)
1, 2, 3, 4, 6, 5, 7, 8, 10, 12, 14, 9, 11, 13, 15, 16, ... .
Explicitly, for 0 ≤ i and 0 ≤ j < 2i we have k2i+ 2j = 2i+ j, and (if i > 0) k2i+ 2j + 1 = 2i+2i –1+ j.
Find a convergent series Σ an such that Σ ak
n diverges.
(c) On the other hand, show that for (kn) as in part (b) above, if Σ ak
n converges then Σ an also converges and Σ ak
n = Σ an.
(It follows that if we write ( jn) for the permutation inverse to (kn), i.e., for each n let jn be the unique integer such that kj
n = n , then the sequence ( jn) has the opposite properties: if Σ an converges then so does Σ aj
n, and Σ aj
n = Σ an, but there exists a divergent series Σ an such that Σ aj
n converges.)
(d) Now let (kn) be an arbitrary sequence of positive integers in which each positive integer occurs once and only once. (This could be described in Rudin’s language as a rearrangement of the sequence of positive integers; or, in different terminology, as a permutation of the positive integers.) For every positive integer N, let us define the mixing number mix ((kn), N ) to be the largest integer M for which there exist M integers n1, n2, ... nM which are alternately ≤ N and > N (i.e., such that if ni ≤ N then ni +1 > N and vice versa), and such that kn
1 < kn
2 < ... < kn
M. In other words, if we color each positive integer m red or blue according to whether it shows up before or after the N th comma in the sequence ...
Answer to True/False question 3.13:0. (a)F. Answers to True/False question 3.14:0. (a) T. (b)T. (c) F.
k1, k2, k3, ... , then mix ((kn), N ) denotes the number of same-color blocks into which the set of all positive integers is divided. Since only N integers are colored red, this number is at most 2N + 1; in particular, it is finite.
If you think through the example you constructed for part (b), you should find that it implicitly used the fact that for the sequence of integers occurring there, mix ((kn), N ) assumed arbitrarily large values, while the positive result of (c) was based on the fact that mix (( jn), N ) never goes higher than 4.
In fact, prove that the following conditions on a rearrangement (kn) of the positive integers are equivalent:
(i) mix ((kn), N ) is bounded as a function of N.
(ii) For every convergent series Σ an of real numbers, the series Σ ak
n also converges.
Show moreover that if these conditions hold, then for every convergent series Σ an of real numbers, Σ ak
n = Σ an.
3.14:4. Rearrangements and the root test. (d : 3)
Find a series which converges by the root test, and a rearrangement of this series for which the root test gives no information.
However, note a reason why, in a situation of this sort, the rearranged series must still converge.
3.14:5. The subsequential limit set of a rearranged convergent series. (d : 1)
Let Σan, α and β be as in the hypothesis of Theorem 3.54, and let Σa′n and (s′n) be as in the conclusion thereof. Assuming the result of 3.2:6(a) (even if you haven’t proven it), deduce that the subsequential limit set of (s′n) is the interval [α,β].
3.14:6. Rearrangements with alternating signs. (d : 5)
Show that if (an) is a non-absolutely convergent series, and α is a real number, then there exists a rearrangement (ak
n) of (an) such that for all even n, ak
n≥0, for all odd n, ak
n≤0, and Σak
n=α. (One could even get the partial sums to have lim sup any α and lim inf any β such that –