Sequences

Section: 
9.1
Date: 
Wednesday, September 29, 2010 - 14:00 - 15:00
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Math 2300 Section 005 – Calculus II – Fall 2010

9.1 Sequences Examples – September 29, 2010

Fill in the empty cells of the following table.

\begin{array}[]{c|c}\textbf{General Term}&\textbf{First 5 Terms}\\<br />
\hline\hline&\\<br />
\hline&\\<br />
s_{n}=\frac{1}{n}&\\<br />
&\\<br />
\hline&\\<br />
s_{n}=n^{2}&\\<br />
&\\<br />
\hline&\\<br />
&1,2,4,8,16,\ldots\\<br />
&\\<br />
\hline&\\<br />
s_{m}=\frac{n}{n+1}&\\<br />
&\\<br />
\hline&\\<br />
&1,-1,1,-1,1,\ldots\\<br />
&\\<br />
\hline&\\<br />
&3.1,3.14,3.141,3.1415\end{array}

We'll introduce the notions of convergent and divergent. You should be able to recognize convergent and divergent sequences by their graphs. We'll also introduce the Fibonacci sequence.

The solutions to the following examples are located below.

  1. Give the first five terms of the sequence

    s_{n}=\frac{n(n+1)}{2}
  2. Give the first five terms of the sequence

    s_{n}=\frac{2n+(-1)^{n}n}{3}
  3. Give a general term for the following sequence. (That is, write the algebraic form for a general term in the sequence.)

    1,2,3,4,5,6,\ldots
  4. Give a general term for the following sequence.

    a_{1}=6,\  a_{2}=2,\  a_{3}=\frac{6}{5},\  a_{4}=\frac{6}{7},\  a_{5}=\frac{2}{3}
  5. Give the first five terms in the Fibonacci sequence, where

    f_{1}=1,\  f_{2}=1,\  f_{n}=f_{{n-1}}+f_{{n-2}}\ (n>2)
  6. Write an algebraic definition for the recursive sequence.

    a_{1}=4,\  a_{2}=9,\  a_{3}=16,\  a_{4}=25.
  7. Does the sequence s_{n}=2+\frac{\pi}{n} converge or diverge?

Solutions

  1. Remember from our work with Riemann sums that this represents the sum of the numbers from 1 to n, and so we could develop the terms in the sequence by simply adding numbers to preceeding sequence elements. Or, we could evaluate the terms in the sequence as usual.

    s_{1}=1,\  s_{2}=3,\  s_{3}=6,\  s_{4}=10,\  s_{5}=15
  2. s_{1}=\frac{1}{3},\  s_{2}=2,\  s_{3}=1,\  s_{4}=4\, s_{5}=\frac{5}{3}
  3. s_{n}=n.

  4. At first, this one is a bit odd. But, notice that a_{1},\, a_{3} and a_{4} all have a 6 in the numerator. Writing all of the terms in the sequence like this (multiplying top and bottom by fractions to maintain equality) gives the following.

    a_{1}=\frac{6}{1},\  a_{2}=\frac{6}{3},\  a_{3}=\frac{6}{5},\  a_{4}=\frac{6}{7},\  a_{5}=\frac{6}{9}

    This now seems like a reasonable quess for an approach, since the numbers in the numerators are all 6 and those in the denominator as odd numbers. So, we can write

    a_{n}=\frac{6}{2n-1}
  5. f_{1}=1,\  f_{2}=1,\  f_{3}=2,\  f_{4}=3,\  f_{5}=5
  6. There are several ways to write this. One way is to recognize that the sequence is a sequence of squares. So we could write

    a_{1}=4,\  a_{n}=\left(\sqrt{a_{{n-1}}}+1\right)^{2}\ \ (n>1).

    For a different formulation, see example 4b in the text.

  7. This sequence is monotone increasing (actually, it is strictly monotone increasing and bounded (since 2<s_{n}\le 2+\pi). Thus, by Theorem 9.1, the sequence converges.

© 2011 Jason B. Hill. All Rights Reserved.