Homework 7

Exercise: Let x\geq 0. Consider the sequence \{a_n\} defined by a_n=x^n. Prove that \{a_n\} converges if and only if x\leq 1. For x\leq 1, determine the limit of \{a_n\}.

Proof. If \{a_n\} converges to a. Clearly a\geq 0. Since a_{n+1}=xa_n, we have a=xa by taking limit on both side. If x=1, obviously a=1. If x\neq 1, the equation a=xa gives us that a=0. Furthermore, if 0\leq x<1, obviously a=\lim{x^n}=0. On the other hand, if x>1, a_n=x^n>1. But we’ve shown that if \{a_n\} converges, its limit is 0. So for x>1, the sequence doesn’t converge.

Exercise: Let a<0 and x_1< be the real numbers. Define a sequence \{x_n\} by
x_{n+1}=\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)\text{ for all }n\in\mathbb{N}

  1. Prove that x_{n+1}\leq x_n for all n\geq 2.
  2. Prove that \{x_n\} converges to x\in\mathbb{R} where x>0 and x^2=a.

Proof.

  1. First, we claim that x_n^2\geq a for all n\geq 2. This is because x_n^2-a=\frac{1}{4}\left(x_{n-1}+a/x_{n-1}\right)^2-a=\frac{1}{4}\left(x_{n-1}-a/x_{n-1}\right)^2\geq 0. Now, we show that x_{n+1}\leq x_n for all n\geq 2. Simply calculate the difference, x_n-x_{n+1}=x_n-(x_n+a/x_n)/2=(x_n^2-a)/2x_n. Easy to see that all x_n‘s are positive. Combining the fact that x_n^2\geq a for all n\geq 2, we know that x_{n+1}\leq x_n.
  2. Since \{x_n\} is decreasing sequence with positive entries, it converges. Suppose the limit is x. Take limit on both sides of the defining equation, we have x=(x+a/x)/2, thus x^2=a. Obviously x is non-negative.

Exercise: For a sequence of real numbers \{a_n\}, denote, by \{s_n\}, the sequence of its arithmetic means as follows
s_n=\frac{1}{n}\sum^n_{k=1}{a+k}=\frac{1}{n}(a_1+a_2+\cdots+a_n)

  1. Prove: If \{a_n\} converges to a, then \{s_n\} converges to a.
  2. Give an example of a divergent sequence \{a_n\} such that \lim_{n\rightarrow\infty}s_n=0
  3. Define the sequence \{b_n\} by b_n=a_{n+1}-a_n for all n\in\mathbb{N}. Suppose \{s_n\} converges to s and \{nb_n\} converges to 0. Prove that \lim_{n\rightarrow\infty}=s. Hint: show that \sum^n_{k=1}{kb_k}=na_{n+1}-\sum^n_{k=1}a_k.

Proof (sketch).

  1. Fix your favorite \epsilon<0 first. There is N such that if n>N, |a_n-a|<\epsilon/2. Denote S=\sum_{k\leq N}(a_k-a). Now for n>N, |s_n-a|=\frac{1}{n}|S+\sum_{N<k\leq n}(a_k-a)|\leq\frac{1}{n}(|S|+(n-N)\epsilon/2). Then try to find M such that if n>M, |s_n-a|<\epsilon. And this shows that \lim s_n=a.
  2. Take a_n=(-1)^n. Then |s_n|\leq 1/n, hence conveges to 0.
  3. Follow the hint, we can prove by induction that \sum_{k=1}^n{kb_k}=na_{n+1}-ns_n. Thus a_{n+1}=s_n+\frac{1}{n}\sum_{k=1}^n{kb_k}. But \{nb_n\} converges to 0, by the previous result, its arithmetic means \{\frac{1}{n}\sum_{k=1}^n{kb_k}\} converges to 0. Hence \{a_n\} converges to the limit of \{s_n\}, which is s.

Exercise: We define the Fibonacci sequence \{F_n\} by
F_0=0, F_1=1 \text{ and } F_{n+1}=F_n+F_{n-1} \text{ for } n\geq 1

  1. Prove that for all n\in\mathbb{N},
    F_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right]
  2. Define the sequence \{a_n\} of real numbers by a_0=1 and a_n=1+1/a_{n-1} for all n\in\mathbb{N}. Prove that for all n\in\mathbb{N}, a_n=1=F_{n+1}/F_{n}.
  3. Prove that \lim_{n\rightarrow\infty}a_n=(1+\sqrt{5})/2.

Proof (sketch).

  1. This can be verified by induction.
  2. Prove by induction. For n=1, a_0=1=1/1=F_2/F_1. Suppose a_{n-1}=F_{n+1}/F_n. Since a_n=1+1/a_{n-1}, we have a_n=1+F_n/F_{n+1}=(F_n+F_{n+1})/F_{n+1}=F_{n+2}/F_{n+1}. This finishes the inductive step.
  3. Denote \alpha=(1+\sqrt{5})/2. Since a_n=F_{n+1}/F_{n}=(\alpha^{n+1}-\alpha^{-n-1})/(\alpha^{n}-\alpha^{-n})=(\alpha-\alpha^{-2n-1})/(1-\alpha^{-2n}), \lim a_n=\alpha.

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