Math 1B  Calculus II  Chapter 8 test (1)  fall ’06           Name__________________________

Show your work for credit.  Write all responses on separate paper.  Don’t abuse a calculator.

 

1.      Show that  can be evaluated using a geometric series of the form, . 
What is the ratio, r  and the leading coefficient, a?   What is the convergent value?
Hint: you’ll need basic properties of exponents like  and .

2.      Use the integral test to determine whether  is convergent or divergent.  
Start by stating what continuous function is and how it meets the conditions of the integral test.

3.      Use the integral test to determine whether the series converges or diverges: . 
Start by stating what continuous function is and how it meets the conditions of the integral test.

4.      Evaluate .  Hint: factor the denominator and find its partial fractions representation…this leads to a telescoping sum.  Write out some terms of the partial fraction representations to see the pattern of cancellation.

5.      Determine whether the following series converges or diverges:  . 
As usual, describe your method in detail.

6.      Use the limit comparison test to determine whether the series converges or diverges: .
Hint: compare with a geometric series.

7.      Show that   is convergent.   What value of N is needed to be sure the approximation  is correct to 20 significant digits?

8.      Write the function  as a power series.  Hint:  .  Use the ratio test to determine the radius of convergence of the series.  What is the interval of convergence?

Math 1B  Calculus II  Chapter 8.1  8.6 test Solutions  fall ’06

1.       
r = 4/9 and the leading coefficient, a = 324.

2.      Since  is a positive valued and decreasing function, we can use the integral test to determine whether  is convergent or divergent.  Substituing  we see that, since the integral is divergent: , the series must also be divergent.

3.      Since  is a positive valued function and, since  on  decreasing function, we can use the integral test to determine whether  converges or diverges.  Substitute ,  to see the integral is convergent: .  So the series is also convergent.

4.      The partial fractions expansion is a simple one:  .  This leads to a telescoping sum: .

5.      The series  converges since  and since , the series must converge.
Alternatively, use the limit comparison test with the convergent p-series, :   so, like the p-series with p = 2,  the series converges.

6.  Using the limit comparison test with the geometric series , we have  must converge since, after a few applications of L’Hopital’s rule,  .

7.   is convergent since it’s alternating and .   To be sure the approximation  is correct to 20 significant digits, we need only require that the magnitude of the first neglected term is . 
Somehow, I don’t think that 1 subtracted from a huge number is going to make a difference, but how can I be more sure?  Well, here’s some calculations in Mathematica that make it seem like the convergence is much faster than that.  Can you confirm the following somehow?

Entering, the command to numerically approximate

      N[Sum[Cos[n*Pi]/Sqrt[n],{n,2,4*^40}],50]

     0.39510135657836962975523408576404450024023745486975
whereas
   N[Sum[(-1)^n/Sqrt[n],{n,2,4*^80}],50]

        0.39510135657836962975273408576404450024026245486975

So that the difference is
   N[Sum[Cos[n*Pi]/Sqrt[n],{n,2,4*^40}],50]-                                                                                   
         N[Sum[Cos[n*Pi]/Sqrt[n],{n,2,4*^80}],50]

 
Which is twice as good as was required.  This makes sense if you consider the nature of the path of this alternating series. Like a pendulum swinging steadily with decreasing arcs of passage, each time it swings, half the length of the swing is a better estimate of the distance from one extreme of the string to the point at equilibrium (about halfway through the swing.)  The distance to the next opposite peak will still cause a rounding error:
       N[Sum[Cos[n*Pi]/Sqrt[n],{n,2,1*^40}],50]-                  |             
         N[Sum[Cos[n*Pi]/Sqrt[n],{n,2,1*^80+1}],50]

 

8.      Writing as a power series,   The ratio test   leaves the endpoints to check.  Since the function is not even defined at x = -1 and at x = 1 the terms are alternating between 1 and -1, the series does not converge at the endpoints of this interval and the interval of convergence is (-1,1).