MATH 1502 - WINTER 2003

EXAM 2 - SOLUTIONS

1.   (20 points)   Evaluate each of the following convergent series:

       (a)   sum((-1)^k/(7^k),k = 2 .. infinity) .    This is a geometric series with x = -1/7   and beginning at k = 2.   

       So sum((-1)^k/(7^k),k = 2 .. infinity)   =   sum((-1)^k/(7^k),k = 0 .. infinity)     -  (1+ 1/7 )  = 1/(1+1/7)-(1-1/7)  =   7/8-6/7   =   1/56     (6 points)

       (b)    sum(2^(k+1)/(5^(k+3)),k = 0 .. infinity)   =    sum([2/(5^3)]*[2^k/(5^k)],k = 0 .. infinity)    =   [2/125]*[sum(2^k/(5^k),k = 0 .. infinity)]    

                                         =    [2/125]*[1/(1-2/5)]    =    [2/125]*[5/3]  = 2/75      (8 points)

       (c)     sum(1/exp(k)-1/exp(k+1),k = 3 .. infinity)   =   1/exp(3) .     (6 points)

        This is a telescoping series with f(k) = 1/exp(k)  .   

        So,   S[n] = 1/exp(3)-1/exp(n+1)   --->   1/exp(3)    because 1/exp(n+1)    --->  0.

2.   (20 points)    Int(1/(1+x^2),x)   =   tan^(-1) (x)  +  C    &       1/(1+x^2) = sum((-1)^k*x^(2*k),k = 0 .. infinity)    (geometric series).

(a)  Find the power series for tan^(-1) (x) by integration;  be sure to find the value of C.   (b)  What is the radius of convergence for the series you obtained in part (a)?

SOLUTION:  (a)    tan^(-1) (x)  +  C  =   Int(1/(1+x^2),x)   

                                                          =    sum(int((-1)^k*x^(2*k),x),k = 0 .. infinity)    =    sum((-1)^k*x^(2*k+1)/(2*k+1),k = 0 .. infinity)  .      (12 points)

To find C, note that tan^(-1) (0) =  0 and the value of the series on the right is 0 when x = 0  (because you have all 0 terms).   So, C = 0.    (3 points)

(b)  The radius of convergence is 1.   We know this because the radius of convergence of the geometric series is 1 and tan^(-1) (x) has the same radius of convergence since it is the indefinite integral of the geometric series.   Or by applying any of the 5 tests.    (5 points)

3.   (20 points)   Find the interval of convergence for the series    sum(2^k*x^k/sqrt(k),k = 1 .. infinity)  .   Be sure to specify the convergence (absolute or conditional) or the divergence at the endpoints.  Be sure to put your answer (the interval of convergence) in the form of an interval.

SOLUTION:   Use the root test (or the ratio test).     (2^k*abs(x)^k/sqrt(k))^(1/k) = 2*abs(x)/sqrt(k^(1/k))   ---->    2*abs(x)   

(because k^(1/k)   --->   1).   By the root test, the series converges absolutely for 2*abs(x)   <   1    

  <===>    abs(x)   <   1/2       <====>     -1/2   <   x    <   1/2  .     (10 points)

Now check the endpoints:

        **     x = -1/2    ===>   sum((-1)^k*2^k*(1/2)^k/sqrt(k),k = 1 .. infinity)    =    sum((-1)^k/sqrt(k),k = 1 .. infinity)    which is an alternating series that converges.  However, the absolute values do not converge, because this is a p-series with p = 1/2 .    So,   at x = -1/2 ,  the series is conditionally convergent.   (5 points)

        **    x = 1/2    ===>   sum(2^k*(1/2)^k/sqrt(k),k = 1 .. infinity)     =     sum(1/sqrt(k),k = 1 .. infinity)    which is a divergent p-series.  (5 points)

So the interval of convergence is   [ -1/2, 1/2 ).   (Since the problem said to put your answer in the form of an interval, count off 4 points if this is not done and all of the above is correct)

4.    (20 points)    Test the convergence or divergence of each of the following series.  Be sure to state explicitly which test (of the 5 tests in the book) you are using; and then just use it.

         (a)      sum(k*4^k/(7^(k+1)),k = 0 .. infinity)                   (b)    sum(1/(k*[ln(k)]^2),k = 2 .. infinity)

SOLUTION:   (a)   Use either the ratio or the root test.     [(k+1)*4^(k+1)/(7^(k+2))]*[7^(k+1)/(k*4^k)]    =    [(k+1)/k]*[4/7]     ---->    4/7    <   1   (because (k+1)/k   ----> 0).   By the ratio test the series converges.   (10 points; count off 3 points if the test used is not explicitly stated)

 (b)   Let    f(x) = 1/(x*[ln(x)]^2)     and  use integration by substitution ( u = ln(x)   and    du = dx/x )  to get:   Int(1/(x*[ln(x)]),x) = -1/ln(x)+C .    So,    Int(1/(x*[ln(x)]^2),x = 2 .. b) = -1/ln(b)+1/ln(2)     --->    1/ln(2)    because 1/ln(b)  ---->  0   as b ----> infinity .   Since the integral converges, the series converges by the integral test.   (10 points; count off 3 points if the test used is not explicitly stated)

5.   (20 points)   Does the following series converge absolutely, conditionally, or does it diverge?   Your answer is one of the three phrases:  converges conditionally, converges absolutely, or diverges.  You still must justify your answer.

        sum((-1)^k/(k+sqrt(k)),k = 1 .. infinity)

SOLUTION:   Test for absolute convergence:   sum(1/(k+sqrt(k)),k = 1 .. infinity)   .    Use the Limit Comparison Test by comparing to a[k] = 1/k .

[1/k]/[1/(k+sqrt(k))]     =      (k+sqrt(k))/k    =    1+1/sqrt(k)    ---->   1  >  0.   By the LCT, the series of absolute values diverges because sum(1/k,k = 1 .. infinity)   diverges.   (10 points)

Test for conditionally convergence.  Use the alternating series test:    1/(k+sqrt(k))   is decreasing and converges to 0.   By the AST, the given series converges.     (10 points)

Therefore, the series converges conditionally.    (Count off 4 points if the type of convergence is not specified)