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CALCULUS 265 Honors
Fall Semester 2000
8:00 to 8:50 MTTF
Carver 0002

Instructor: Hentzel
Office: 432 Carver
Office Hours: 9:00-10:00 MTTF

E-mail  hentzel@iastate.edu
phone  294-8141
Website

http://www.math.iastate.edu/hentzel/honors.265

Monday, November 13:    Section  8.3
p578:5,9,15,23,29,30,31,32,33

Monday, November 13
p579:34,35,36,37,38,41,42,47,48,49,50

previous assignment
p571:7,11,12,19
p572:33,37

Main idea:  Bigger than big is BIG.
            Smaller than small is SMALL.

Key Words: Direct Comparison Test, Limit Comparison test. 

Goal: Learn how to tell if a sequence converges. 

p578:5,9,15,23,29,30,31,32,33

              1 
#5    SUM   ------
             n-1

n-1 <= 	n

  1       1
----  >= ---
 n-1      n


Sum 1/n is the harmonic series which diverges.

Sum 1/(n-1) diverges by direct comparison test.


         ln n
#9  SUM -----
          n+1


ln n > 1   eventually (n>=3)

ln n        1
-----   >= -----
n+1        n+1


Sum 1/(n+1) is the harmonic series and diverges.

Sum ln n/(n+1) diverges by direct comparison test.




            n
#15  SUM  -----
          n^2+1

Let bn = 1/n
                          n         n
Limit   an/bn = Limit  -------- * -----   = 1
                       n^2 + 1      1


Sum 1/n is the harmonic series and diverges.

Sum 1/(n^2+1) diverges by limit comparison test.


p578:5,9,15,23,29,30,31,32,33
 
               1
#23  SUM    ----------
            n Sqrt[n^2 + 1]

Let bn = 1/n^2
                       1                n^2               n^2
Limit an/bn = Limit  -------------- * ------ = Limit -------------  = 1
                    n Sqrt[n^2 + 1]     1            n Sqrt[n^2+1]


Sum bn = Sum 1/n^2 is a p series, p=2 and converges.

       1
Sum -------------  converges by limit comparison test.
    n Sqrt[n^2+1]


#29  Sum  Sqrt[n]/n

This is a p-series, p=1/2 and diverges.


#30  Sum 5(-1/5)^n

This is a geometric series, r = -1/5 and converges.

#31  Sum 3/(3^n + 2)

Let bn = 1/3^n
                       3      3^n
Limit an/bn = Limit ------ * ------ =  3
                    3^n+2     1  

Sum 1/3^n is a geometric series, r=1/3 and converges.

Sum 3/(3^n+2) converges by limit comparison test.


               1
#32  Sum ----------------
         3 n^2 - 2 n - 15


Let bn = 1/n^2

           1                n^2
Limit ---------------- * -----------  = 1/3
      3 n^2 -2 n -15        1

Sum bn is a p-series, p = 2 and converges.

Sum 1/(3n^2 -2n -15) converges by limit comparison test.

               n
#33  Sum --------------
           2 n + 3


Limit an = 1/2 so Sum n/(2n+3) diverges by nth term test.


#34  Sum (1/(n+1) - 1/(n+2))

This is a telescoping series and converges to the first term.

          n
#35  Sum -----
        (n^2+1)^2

          n               -1/2        |n=Infinity
Int   -----------  =    ---------     |             and converges.
       (n^2+1)^2        n^2 + 1       |n=1

              n           
Derivative ---------  =
           (n^2+1)^2       

               2
        1 - 3 n
Out[2]= ---------
              2 3
        (1 + n )

This is decreasing for n >= 1.

        3
Sum -----------
     n(n+3)

   a           b            3
------  +  ---------  = -----------
   n         n+3          n(n+3)


na + 3a + bn = 3

a+b = 0  a = 1

         3               1          12
Sum ------------  =  ---------- - --------
       n(n+1)           n          n+3


This is a telescoping series and converges to the sum of the first

two terms.