Math 312, Intro. to Real Analysis: Homework #4 Solutions

Stephen G. Simpson

Monday, March 2, 2009

The assignment consists of Exercises 10.6, 10.8, 10.10, 11.2, 11.3, 11.6, 11.10, 12.2, 12.12 in the Ross textbook. Each problem counts 10 points.

10.6. (a) Let (s_{n}) be a sequence of real numbers. Then, whenever m is less than n we have

s_{m} − s_{n} = (s_{m} − s_{m+1}) + (s_{m+1} − s_{m+2}) + ⋯ + (s_{n−1} − s_{n}),

hence by the Triangle Inequality

|s_{m} − s_{n}| ≤ |s_{m} − s_{m+1}| + |s_{m+1} − s_{m+2}| + ⋯ + |s_{n−1} − s_{n}|.

Now, if our sequence is such that |s_{k} − s_{k+1}| ≤ 1/2^{k} for all k, then our inequality becomes

|s_{m} − s_{n}| ≤ \frac{1}{2^{m}} + \frac{1}{2^{m+1}} + ⋯ + \frac{1}{2^{n−1}}

and the right-hand side is the beginning of the geometrical series ∑_{k=m}^{∞} 1/2^{k} = 1/2^{m−1}. Thus we see that |s_{m} − s_{n}| ≤ 1/2^{m−1} for all m and all n > m. Thus (s_{n}) is a Cauchy sequence. Therefore, by one of our theorems, (s_{n}) is a convergent sequence.

(b) If we assume only that |s_{k} − s_{k+1}| < 1/k for all k, we cannot conclude that (s_{n}) is a Cauchy sequence. An example illustrating this is obtained by letting s_{n} = the nth partial sum of the harmonic series,

s_{n} = H_{n} = 1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n}.

For this example we have

|s_{n} − s_{n+1}| = \frac{1}{n + 1} < \frac{1}{n}

but

lim s_{n} = 1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{n} + ⋯ = ∑_{n=1}^{∞} \frac{1}{n} = +∞ (harmonic series)

so our sequence (s_{n}) is divergent.

10.8. Assume that (s_{n}) is a nondecreasing sequence of real numbers. Let σ_{n} be the average of the first n numbers in our given sequence:

σ_{n} = \frac{s_{1} + ⋯ + s_{n}}{n} .

We claim that the sequence (σ_{n}) is again nondecreasing. To see this, note that s_{1} ≤ ⋯ ≤ s_{n} ≤ s_{n+1}, hence s_{1} + ⋯ + s_{n} ≤ ns_{n+1}. Adding n(s_{1} + ⋯ + s_{n}) to both sides, we obtain

n(s_{1} + ⋯ + s_{n}) + (s_{1} + ⋯ + s_{n}) ≤ n(s_{1} + ⋯ + s_{n}) + ns_{n+1},

or, in other words,

(n + 1)(s_{1} + ⋯ + s_{n}) ≤ n(s_{1} + ⋯ + s_{n} + s_{n+1}).

Dividing both sides by n(n + 1), we obtain

\frac{s_{1} + ⋯ + s_{n}}{n} ≤ \frac{s_{1} + ⋯ + s_{n} + s_{n+1}}{n + 1}

or, in other words, σ_{n} ≤ σ_{n+1} for all n. This proves our claim.

The conclusion can be restated as follows:

If a sequence is nondecreasing, then its sequence of averages is nondecreasing.

Note: The textbook assumes in addition that s_{n} is positive for all n. However, this assumption is not needed in order to obtain the conclusion.

10.10. Define a sequence (s_{n}) inductively by letting s_{1} = 1 and s_{n+1} = (s_{n} +1)/3 for all n.

(a) The first few terms of the sequence are s_{1} = 1 and s_{2} = 2/3 and s_{3} = 5/9 and s_{4} = 14/27.

(b) Obviously s_{1} > 1/2. If s_{n} > 1/2, it follows that s_{n} + 1 > 3/2, hence s_{n+1} = (s_{n} + 1)/3 > (3/2)/3, hence s_{n} > 1/2. Thus by induction on n we see that s_{n} > 1/2 for all n.

(c) Since s_{n} > 1/2, it follows that

s_{n} − s_{n+1} = s_{n} − \frac{s_{n} + 1}{3} = \frac{2}{3}s_{n} − \frac{1}{3} > \frac{2}{3} · \frac{1}{2} − \frac{1}{3} = 0

so s_{n} > s_{n+1} for all n, i.e., (s_{n}) is a decreasing sequence.

(d) Since (s_{n}) is a decreasing sequence which is bounded below (the bound is 1/2), it follows by one of our theorems that (s_{n}) is convergent. Let α = lim s_{n}. Taking the limit of both sides of the equation s_{n+1} = (s_{n} + 1)/3, we obtain α = (α + 1)/3. Solving for α we obtain α = 1/2. Thus lim s_{n} = 1/2.

11.2. Let

a_{n} = (−1)^{n},  b_{n} = \frac{1}{n},  c_{n} = n^{2},  d_{n} = \frac{6n + 4}{7n − 3}.

Note that the sequence (a_{n}) looks like −1, 1, −1, 1, . . ..

(a) A monotone subsequence of (a_{n})_{n} is 1, 1, 1, . . .. We can describe this subsequence as (a_{n_{k}})_{k} where n_{k} = 2k. Thus a_{n_{k}} = (−1)^{2k} = 1 for all k. Alternatively, using n instead of k as the index, we can describe our subsequence as (a_{2n}). The sequences (b_{n}), (c_{n}), and (d_{n}) are already monotone. (Note that any sequence is considered to be a subsequence of itself.) To see that (d_{n}) is monotone, use algebra to show directly that

\frac{6n + 4}{7n − 3} > \frac{6(n + 1) + 4}{7(n + 1) − 3}

for all n.

(b) The subsequential limits of our sequences are −1 and 1 for (a_{n}), 0 for (b_{n}), +∞ for (c_{n}), and 6/7 for (d_{n}).

(c) lim sup a_{n} = 1, lim inf a_{n} = −1. lim sup b_{n} = lim inf b_{n} = lim b_{n} = 0. lim sup c_{n} = lim inf c_{n} = lim c_{n} = +∞. lim sup d_{n} = lim inf d_{n} = lim d_{n} = 6/7.

(d) (a_{n}) is divergent and its limit is undefined. (b_{n}) and (d_{n}) are convergent to 0 and 6/7 respectively. (c_{n}) is divergent to +∞.

(e) (a_{n}), (b_{n}), and (d_{n}) are bounded. (c_{n}) is unbounded.

11.3. Let

s_{n} = cos \frac{nπ}{3} ,  t_{n} = \frac{3}{4n + 1},  u_{n} = (−\frac{1}{2})^{n} ,  v_{n} = (−1)^{n} + \frac{1}{n}.

Note that the sequence (s_{n}) looks like

\frac{1}{2}, −\frac{1}{2}, −1, −\frac{1}{2}, \frac{1}{2}, 1, \frac{1}{2}, −\frac{1}{2}, −1, ….

(a) A monotone sbsequence of (s_{n}) is −1, −1, −1, …, which can be described as (s_{6n+3}), i.e., (cos(2n + 1)π). The sequence (t_{n}) is already monotone. A monotone subsequence of (u_{n}) is (u_{2n}), i.e., (1/2^{2n}), which looks like 1/4, 1/16, 1/64, . . .. A monotone subsequence of (v_{n}) is (v_{2n}), i.e., (1 + 1/2n), which looks like 3/2, 5/4, 7/6, . . ..

(b) The subsequential limits are ±1, ±1/2 for (s_{n}), 0 for (t_{n}), 0 for (u_{n}), ±1 for (v_{n}).

(c) (s_{n}) and (v_{n}) are divergent and their limits are undefined. (t_{n}) and (u_{n}) are convergent to 0.

(d) (s_{n}), (t_{n}), (u_{n}), and (v_{n}) are bounded.

11.6. Assume that (c_{n}) is a subsequence of (b_{n}) and (b_{n}) is a subsequence of (a_{n}). By equation (3) on page 64, there are functions f(n) and g(n) such that b_{n} = a_{f(n)} and c_{n} = b_{g(n)} for all n. It follows that c_{n} = a_{f(g(n))} for all n. Thus (c_{n}) is a subsequence of (a_{n}) via the composite function f ∘ g.

11.10. We are given the sequence 1, 1/2, 1, 1, 1/2, 1/3, 1/4, 1/3, 1/2, 1, 1, 1/2, . . .. The subsequential limits of this sequence are 1, 1/2, 1/3, …, 1/n, … and 0. The lim sup is 1 and the lim inf is 0.

12.2. By definition of limits, lim s_{n} = 0 means that for all ϵ > 0 we can find N so large that |s_{n} − 0| < ϵ for all n > N. In other words,

for all ϵ > 0 we can find N so large that |s_{n}| < ϵ for all n > N.

But then, by the definition of lim sup, this condition is precisely equivalent to saying that lim sup |s_{n}| ≤ 0. Moreover, |s_{n}| ≥ 0 automatically, hence lim sup |s_{n}| ≥ 0 automatically, for all sequences. Thus, our condition above is equivalent to saying that lim sup |s_{n}| = 0.

We have now proved that lim s_{n} = 0 if and only if lim sup |s_{n}| = 0.

12.12. Assume that (s_{n}) is a sequence of real numbers. As in Exercise 10.8, let σ_{n} be the average of s_{1}, …, s_{n}.

(a) We shall prove the chain of inequalities

lim inf s_{n} ≤ lim inf σ_{n} ≤ lim sup σ_{n} ≤ lim sup s_{n}.  (1)

The middle inequality lim inf σ_{n} ≤ lim sup σ_{n} is obvious and does not require proof. We prove the right-hand inequality lim sup σ_{n} ≤ lim sup s_{n}. We omit the proof of the left-hand inequality lim inf s_{n} ≤ lim inf σ_{n}, which is similar. In order to prove lim sup σ_{n} ≤ lim sup s_{n}, we follow the hint given in the textbook. First, given M and N such that M > N, we claim that

sup_{n>M} σ_{n} ≤ \frac{s_{1} + ⋯ + s_{N}}{M} + sup_{k>N} s_{k}.  (2)

To prove (2), it suffices to prove

σ_{n} ≤ \frac{s_{1} + ⋯ + s_{N}}{M} + sup_{k>N} s_{k}  (3)

for each individual n > M. Since n > M > N, we can break σ_{n} into two parts:

σ_{n} = \frac{s_{1} + ⋯ + s_{n}}{n} = \frac{s_{1} + ⋯ + s_{N}}{n} + \frac{s_{N+1} + ⋯ + s_{n}}{n} . (4)

Since n > M, the first part of the right-hand side of (4) is < the first part of the right-hand side of (3). And obviously, the second part of the right-hand side of (4) is ≤ the second part of the right-hand side of (3). Combining these two observations, we see that (3) holds. From this it follows immediately that (2) holds. Now, fixing N and taking the limit of (2) as M goes to infinity, we see that

lim sup σ_{n} ≤ sup_{k>N} s_{k}  (5)

for all N. Finally, taking the limit of (5) as N goes to infinity, we see that

lim sup σ_{n} ≤ lim sup s_{k}

and this completes the proof.

(b) In the case when lim s_{n} exists, we have lim s_{n} = lim inf s_{n} = lim sup s_{n}, so (1) becomes

lim inf s_{n} = lim inf σ_{n} = lim sup σ_{n} = lim sup s_{n} = lim s_{n}.

From this it follows that lim σ_{n} exists and is equal to lim s_{n}. This conclusion can be restated as follows:

If a sequence is convergent, then its sequence of averages is convergent.