Problem Set #4 Part C Official Solutions

Inna Zakharevich

October 27, 2003

(15) 1.

(2) (1) Note that

y_{n} = sup_{m≥n} x_{m} = max{x_{n}, sup_{m≥n+1} x_{m}} = max{x_{n}, y_{n+1}} ≥ y_{n+1}.

Thus the sequence {y_{n}} is nonincreasing. In addition, we know that there exists M such that x_{n} ≥ M for all n. Since y_{n} ≥ x_{n} by definition, we know that y_{n} ≥ M for all n. Thus {y_{n}} is a nonincreasing sequence bounded from below.

(2) (2) Define z_{n} = inf_{m≥n} x_{m}. Let lim inf x_{n} = lim z_{n}.

(5) (3) Note that

y_{n} = sup_{m≥n} x_{n} ≥ inf_{m≥n} x_{n} = z_{n}.

Note that if we have a convergent sequence where all elements are nonnegative, the limit of the sequence must also be nonnegative. Thus the limit of the sequence {y_{n} − z_{n}} is nonnegative, so lim sup x_{n} ≥ lim inf x_{n}. Now suppose that lim sup x_{n} = lim inf x_{n}. Fix ϵ > 0. Then there will be an N such that for all n ≥ N we have y_{n} − z_{n} < ϵ. This means that for all m_{1}, m_{2} > N we have |x_{m_{1}} − x_{m_{2}}| < ϵ, so {x_{n}}_{n≥N} is a Cauchy sequence, so it converges, and so has a limit. Note that the limit of any sequence is between lim sup and lim inf of that sequence, so if they are equal the sequence must converge to the lim sup. Now suppose that {x_{n}} is a convergent sequence, with limit l. Fix ϵ > 0. Then there will exist an N such that for all n ≥ N |x_{n}−l| < ϵ/2. Thus for any m_{1}, m_{2} ≥ N we will have |x_{n} − x_{m}| ≤ |x_{n} − l| + |l − x_{m}| < 2ϵ. Thus y_{N} − z_{N} ≤ ϵ. Thus lim{y_{n} − z_{n}} = 0, so lim sup x_{n} = lim inf x_{n}. Thus since lim sup x_{n} = lim inf x_{n} we see that lim sup x_{n} = lim x_{n}, so we are done.

(4) (4) A(x) will be bounded because {x_{n}} is bounded. Then A(x) is compact, so it will contain its largest element.

Fix ϵ > 0. Let {x_{ϕ(n)}} be a subsequence of {x_{n}} converging to L, the largest element of A(x), and suppose that l < L. Let e = (L − l)/2. Then we can assume without loss of generality that x_{ϕ(n)} > l + e for all n. Thus y_{n} = sup_{m≥n} x_{n} ≥ x_{ϕ(n)} > l+e for all n, so l = lim sup x_{n} = lim y_{n} ≥ l+e > l which is a contradiction. Thus l is the largest element of A(x). Note that the number of n ∈ N such that x_{n} > l + ϵ is finite, since if it were not we would have a subsequence {x_{ϕ(n)}} such that each element will be larger than l + ϵ, so it will have an accumulation point larger than l + ϵ, which contradicts the fact that l is the largest element.

(2) (5) lim inf x_{n} will be the smallest element of A(x), s. For every ϵ > 0 the set {x_{n} : x_{n} < s − ϵ} is finite.

(10) 2. (Sarah Eggleston) First we will show that the sequence is bounded above, so that it is a bounded sequence. Note that

nα_{n} ≤ (n − 1)α_{n−1} + α_{1} ≤ (n − 2)α_{n−2} + 2α_{1} ⋯ ≤ nα_{n}

so α_{n} ≤ α_{1}. Now let l = lim inf α_{n}. We will show that there exists an N such that for all n > N, |l − α_{n}| < δ. Fix k such that α_{k} < l + δ/2 (which exists by the definition of l).

Let m > kα_{1}/δ, and define M ∈ N to be such that m is between Mk and (M + 1)k. Then note that α_{Mk} ≤ α_{k} (by the same method we used above). Then

α_{m} ≤ \frac{Mk}{m} α_{Mk} + \frac{m − Mk}{m} α_{m−Mk} ≤ \frac{Mk}{m} α_{k} + \frac{m − Mk}{m} α_{1}.

Then from our definition of M, we know that (m − Mk)/m < k/m. So

α_{m} ≤ α_{k} + (kα_{1})/m < l + δ/2 + δ/2 = l + δ.

Thus for all m > kα_{1}/δ, α_{m} < l + δ. We know that there are finitely many a such that α_{a} < l. Let A be the maximum such a, and let N = max{kα_{1}/delta, A}. Then for n > N, we know that |α_{n} − l| < δ. Thus l is the limit of {α_{n}}, so we are done.

(15) 3.

(5) (1) Let s_{n} = ∑_{i=1}^{n} a_{n} and let s′_{n} = ∑_{i=1}^{n} a_{ϕ(n)}. Let A_{n} = ∑_{i=1}^{n} |a_{n}|. Then we want to show that lim s′_{n} = lim s_{n}. Define f(n) = max{ϕ(m) : m ≤ n}. We know that there exists an N such that for all n > N we have |A_{n} − lim A_{n}| < ϵ. Then for m ≥ f(N) we have

|s′_{m} − s_{m}| = |  ∑_{ϕ(m)>N or m>f(N)}(−1)^{∗}a_{m}| ≤ ∑_{ϕ(m)>N or m<f(N)} |a_{m}| < ∑_{m=N}^{∞} |a_{m}| < ϵ.

Thus lim(s_{n} − s′_{n}) = 0, so lim s_{n} = lim s′_{n}.

(5) 2. Suppose that either P or N is finite. Without loss of generality, N is finite. Then ∑_{n∈N} a_{n} = −M for some M. Since ∑ a_{n} is convergent, we know that ∑_{n>max N} a_{n} converges. It is equal to ∑_{n>max N} |a_{n}|. But then ∑ |a_{n}| = ∑_{n<max N} |a_{N}| + ∑_{n>max N} |a_{n}| which is a finite sum plus a convergent sum, so it converges. This is a contradiction. So both P and N are infinite. First note that if one of ∑_{p∈P} a_{p} or ∑_{n∈N} diverges then they both must, since if only one did then the total sum would diverge, a contradiction. Now suppose that both of these sums converge to A and B, respectively. Note that each of these sums must converge absolutely. Then

∑_{n=1}^{k} |a_{n}| = ∑_{n∈N,n≤k} |a_{n}| + ∑_{n∈P,n≤k} |a_{n}| = | ∑_{n∈N,n≤k} a_{n}| + | ∑_{n∈P,n≤k} a_{n}| ≤ |A| + |B|

so ∑ a_{n} is absolutely convergent, a contradiction. So each of ∑_{n∈N} a_{n} and ∑_{n∈P} a_{n} must diverge.

(5) (3-4) Since ∑ a_{n} is convergent, we know that |a_{n}| → 0 as n → ∞. Let N be such that |a_{n}| < ϵ for all n > N, and that for some k_{1}, k_{2} < N ∑_{i=1}^{k_{1}} a_{ϕ(i)} > z and ∑_{i=1}^{k_{2}} a_{phi(i)} < z. Let M be such that ϕ(m) > N for all m > M. Then we know that for all m > M, if ∑_{i=1}^{m} a_{ϕ(i)} > z then ∑_{i=1}^{m} a_{ϕ(i)} − z < |a_{ϕ(m)}| < ϵ, so we see that lim sup ∑_{i=1}^{m} a_{ϕ(i)} ≤ z. With a similar argument, we see that lim inf ∑_{i=1}^{n} a_{ϕ(i)} ≥ z. Thus we see that lim sup ∑_{i=1}^{n} a_{ϕ(i)} = lim inf ∑_{i=1}^{n} a_{ϕ(i)} = z so the sum converges to z.