# Homework 5

Exercise: Use the definition of convergence to determine the following limits in \mathbb{R}:

1. \lim_{n\rightarrow\infty}\frac{3n+1}{n+5}
2. \lim_{n\rightarrow\infty}(\sqrt{n^2+n}-n)

Proof.

1. \lim_{n\rightarrow\infty}\frac{3n+1}{n+5}=3
2. \lim_{n\rightarrow\infty}(\sqrt{n^2+n}-n)=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{2}

Exercise:

1. Give an example of a convergent sequence \{s_n\} of positive real numbers such that \lim_{n\rightarrow\infty}\frac{s_{n+1}}{s_n}=1
2. Give an example of a divergent sequence \{s_n\} of positive real numbers such that \lim_{n\rightarrow\infty}\frac{s_{n+1}}{s_n}=1

Proof.

1. s_n=1
2. s_n=n

Exercise:

1. Give an example of a convergent sequence \{s_n\} of real numbers such that the set \{s_n:n\in\mathbb{N}\} has exactly one limit point.
2. Give an example of a convergent sequence \{s_n\} of real numbers such that the set \{s_n:n\in\mathbb{N}\} has no limit point.
3. Prove: If a sequence \{s_n\} of real numbers converges, then the set \{s_n:n\in\mathbb{N}\} has at most one limit point.

Proof.

1. s_n=1/n
2. s_n=1
3. Suppose \lim_{n\rightarrow\infty}s_n=s. Enough to show that if there is a limit point t for the set \{s_n:n\in\mathbb{N}\}, then s=t. Deny. Then pick any positive real number \epsilon<|s-t|/2. There is an integer N such that n\geq N implies |s-s_n|<\epsilon. Observe that now t is still the limit point of the set \{s_n:n\geq N\}, because chopping off finitely many points doesn’t change the limit point. However, by the choice of \epsilon and N, for all n\geq N, |t-s_n|>|s-t|/2>0. A contradiction.

Exercise: Suppose X\neq\emptyset is equipped with the discrete metric. Characterize all convergent sequences in X.

Proof.
We say a sequence \{s_n:n\in\mathbb{N}\} in X is eventually constant if and only if there are N\in\mathbb{N} and s\in\ X such that for all n>N, s_n=s.
We claim that all eventually constant sequences are all convergent sequences in X. Easy to see they are convergent. Left to see that every convergent sequence is eventually constant. Suppose \{s_n:n\in\mathbb{N}\} convergences to s. By definition of convergence, there is an integer N such that n>N implies that d(s_n,s)<1/2, which, indeed, implies s_n=s because of the discrete metric.

Exercise:

1. Prove: If a sequence \{s_n\}\subset\mathbb{R} converges in \mathbb{R} then the sequence \{|s_n|\} converges in \mathbb{R}.
2. Give an example of a sequence \{s_n\}\subset\mathbb{R} such that \{|s_n|\} converges but \{s_n\} does not converge.
3. For a sequence \{s_n\}\subset\mathbb{R} we define the sequences \{s_n^+\} and \{s_n^-\} where s_n^+:=\max\{s_n, 0\} and s_n^-:=\min\{s_n,0\} for all n\in\mathbb{N}. Prove: the sequence \{s_n\} converges if and only if \{s_n^+\} and \{s_n^-\} converge.

Proof.

1. Suppose \{s_n\} converges to s. Then for every \epsilon > 0, there is an integer N such that n\geq N implies |s_n-s| < \epsilon. Notice that ||s_n|-|s||\leq |s_n-s| < \epsilon. Hence \{|s_n|\} converges to |s|.
2. For example, s_n=(-1)^n.
3. Suppose \{s_n\} converges. Notice that s_n^+=(s_n+|s_n|)/2 and s_n^-=(s_n-|s_n|)/2. By (a) and theorem 3.4, we know \{s_n^+\} and \{s_n^-\} also converge. On the other hand, if \{s_n^+\} and \{s_n^-\} converge, \{s_n\} converges since s_n=s_n^++s_n^-.

1. markwongsk says:
1. Roy says: