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Homework 3

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Exercise 8.6(a)

Let (sn)\p{s_n} be a sequence in R\R. Prove that lim⁡n→∞sn=0\lim_{n\to\infty} s_n = 0 if and only if lim⁡n→∞∣sn∣=0\lim_{n\to\infty} \abs{s_n} = 0.

Solution.

"  ⟹  \implies"

Let ε>0\varepsilon > 0. Since lim⁡n→∞sn=0\lim_{n\to\infty} s_n = 0, there exists N∈RN \in \R such that if n>Nn > N, then ∣sn−0∣<ε\abs{s_n - 0} < \varepsilon. This same NN will work for ∣sn∣\abs{s_n}. Since ∣sn∣≥0\abs{s_n} \geq 0, we have

∣∣sn∣−0∣=∣sn−0∣<ε\abs{\abs{s_n} - 0} = \abs{s_n - 0} < \varepsilon

for n>Nn > N.

Common Mistakes

One mistake I saw a few times was something like this:

There are two cases: ∣sn∣=sn\abs{s_n} = s_n or ∣sn∣=−sn\abs{s_n} = -s_n. In either case, we have lim⁡n→∞sn=lim⁡n→∞(−sn)=0\lim_{n\to\infty} s_n = \lim_{n\to\infty} \p{-s_n} = 0. Therefore, lim⁡n→∞∣sn∣=0\lim_{n\to\infty} \abs{s_n} = 0.

The issue with this is that this does not exist all the cases. While it's true that for each nn, we have ∣sn∣=sn\abs{s_n} = s_n or ∣sn∣=−sn\abs{s_n} = -s_n, the issue is that the sign of sns_n could change with nn.

In other words, for this argument to work, you would need (∣sn∣)=(sn)\p{\abs{s_n}} = \p{s_n} or (∣sn∣)=(−sn)\p{\abs{s_n}} = \p{-s_n}, but this doesn't have to happen, e.g., sn=(−1)ns_n = \p{-1}^n.

Exercise 8.7(b)

Show that sn=(−1)nns_n = \p{-1}^n n does not converge.

Solution.

Note: There are a lot of ways to do this problem. This is the one that first came to my mind.

Suppose for the sake of contradiction that (sn)\p{s_n} does converge to some s∈Rs \in \R. Let ε=12\varepsilon = \frac{1}{2}. Then there exists N∈RN \in \R such that if n>Nn > N, then

∣sn−s∣<12  ⟹  s−12<(−1)nn<s+12.\abs{s_n - s} < \frac{1}{2} \implies s - \frac{1}{2} < \p{-1}^n n < s + \frac{1}{2}.

In particular, 2n>N2n > N, so

s−12<2n<s+12  ⟹  2n−12<s.s - \frac{1}{2} < 2n < s + \frac{1}{2} \implies 2n - \frac{1}{2} < s.

On the other hand, we have 2n+1>N2n + 1 > N, so

s−12<−(2n+1)<s+12  ⟹  s<−(2n+12).s - \frac{1}{2} < -\p{2n + 1} < s + \frac{1}{2} \implies s < -\p{2n + \frac{1}{2}}.

But this implies

0≤2n−12<s<−(2n+12)≤0,0 \leq 2n - \frac{1}{2} < s < -\p{2n + \frac{1}{2}} \leq 0,

which is a contradiction.

Exercise 9.4

Let s1=1s_1 = 1 and for n≥1n \geq 1 let sn+1=sn+1s_{n+1} = \sqrt{s_n + 1}.

  1. List the first four terms of (sn)\p{s_n}.
  2. It turns out that (sn)\p{s_n} converges. Assume this fact and prove the limit is 12(1+5)\frac{1}{2} \p{1 + \sqrt{5}}.
Solution.

  1. 1,2,2+1,2+1+11, \sqrt{2}, \sqrt{\sqrt{2} + 1}, \sqrt{\sqrt{\sqrt{2} + 1} + 1}

  2. Assuming (sn)\p{s_n} converges, we use Example 5 from the book (which I showed you in discussion), to get

    lim⁡n→∞sn+1=s+1.\lim_{n\to\infty} \sqrt{s_n + 1} = \sqrt{s + 1}.

    On the other hand, it's not too hard to show that lim⁡n→∞sn+1=s\lim_{n\to\infty} s_{n+1} = s also. Thus, we get

    s=lim⁡n→∞sn+1lim⁡n→∞sn+1=s+1.s = \lim_{n\to\infty} s_{n+1} \lim_{n\to\infty} \sqrt{s_n + 1} = \sqrt{s + 1}.

    This implies s2−s−1=0s^2 - s - 1 = 0, so

    s∈{1+52,1−52}.s \in \set{\frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}}.

    You can show by induction that sn≥0s_n \geq 0 for all nn, which implies s≥0s \geq 0 as well (why?). In particular, this eliminates 1−52\frac{1 - \sqrt{5}}{2}, so

    s=1+52.s = \frac{1 + \sqrt{5}}{2}.
Exercise 1.

Show that if n0∈Nn_0 \in \N and lim⁡n→∞sn=s\lim_{n\to\infty} s_n = s, then lim⁡n→∞sn+n0=s\lim_{n\to\infty} s_{n+n_0} = s as well. (Hint: Take the NN you get from (sn)\p{s_n} and use N′=N−n0N' = N - n_0.)
Exercise 2.

  1. Show that if sn≥0s_n \geq 0 for all n≥1n \geq 1 and lim⁡n→∞sn=s\lim_{n\to\infty} s_n = s, then s≥0s \geq 0 as well.

  2. Give a counter-example to show that this statement is false with strict inequalities, i.e., to show that the statement

    If sn>0s_n > 0 for all n≥1n \geq 1 and lim⁡n→∞sn=s\lim_{n\to\infty} s_n = s, then s>0s > 0 as well.

    is false.