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  • STA256 Lecture 01

Sequence

• A sequence is a set of ordered objects.
• Increasing Sequence
  • Where the terms are getting bigger
• Non-decreasing Sequences
  • Non-decreasing sequences can have terms which are bigger or stay the same as it progresses.
  • Different from Increasing sequences which must get bigger and bigger.
  • Where $C_n\subset C_{n+1}$ for all $n$
  • So $\underset{n\to \mathrm{\infty }}{lim}{C}_n=\bigcup _{n=1}^{\mathrm{\infty }}{C}_n$
• Show that this sequence of sets is increasing, find $\underset{n\to \mathrm{\infty }}{lim}{C}_k$
• $C_k=\{x:\frac{1}{k}\le x\le 3-\frac{1}{k}\}=1,2,3$
  • $k=1:c_1=\{x:\frac{1}{1}\le x\le 2\}$
  • $k=2:c_2=\{x:\frac{1}{2}\le x\le \frac{5}{2}\}$
  • $k=3:c_3=\{x:\frac{1}{3}\le x\le \frac{8}{3}\}$
  • Notice that the lower bound gets lower and lower and the higher bound gets higher and higher.
  • $\underset{k\to \mathrm{\infty }}{lim}3-\frac{1}{k}=3$
  • $\underset{k\to \mathrm{\infty }}{lim}\frac{1}{k}=0$
  • $\underset{n\to \mathrm{\infty }}{lim}{C}_n=\{x:0<x<3\}$
  • Base Case $k=1$
  • Induction Hypothesis:
    • $C_n\subset C_{n+1}$
    • $\{x:\frac{1}{n}\le x\le 3-\frac{1}{n}\}\subset \{x:\frac{1}{n+1}\le x\le 3-\frac{1}{n+1}\}$
    • $\{x:\frac{1}{n}\le x\le 3-\frac{1}{n}\}\subset \{x:\frac{1}{n+1}\le x\le 3-\frac{1}{n+1}\}$
    • Show that $\frac{1}{n}>\frac{1}{n+1}$
    • Done simple, this shows that the lower bound gets smaller, which means it allows more elements in the set $C_{n+1}$
    • Show that $3-\frac{1}{n}<3-\frac{1}{n+1}$.
      • $3-\frac{1}{n}<3-\frac{1}{n+1}$.
      • $\frac{-\frac{1}{n}<-\frac{1}{n+1}}{-1}$.
      • $\frac{1}{n}>\frac{1}{n+1}$
      • So this shows the case as before, so now we know that the lower bound gets lower and the upper bound get higher.
      • If this is the case, this must mean that the range gets wider, which means that it allows more elements in, so $C_n\subset C_{n+1}$