Page 272 - Calculus Workbook For Dummies

256 Part IV: Integration and Infinite Series

3
e ! 1 diverges.
n
n 1
=
1
With the p-series rule, you can now solve problem 2. ! , called the harmonic series
n
1 1 1 1 1
(1 + + + + + ... + ), is probably the most important p-series. Because p = 1, the
2 3 4 5 n
p-series rule tells you that the harmonic series diverges.
4 2 4 3 4 4 4 n
f 1 + + + + ... + diverges.
2 3 4 n
This may not look like a p-series, but you can’t always judge a book by its cover.
2 / 1 4 3 / 1 4 4 / 1 4 n / 1 4
=
1. Rewrite the terms with exponents instead of roots: 1 + + + + ... + .
2 3 4 n
2. Use ordinary laws of exponents to move each numerator to the denominator.
1 1 1 1
= 1 + / 3 4 + / 3 4 + / 3 4 + ... + / 3 4
2 3 4 n
3
3. Apply the p-series rule. You’ve got a p-series with p = , so this series diverges.
4
n
3
g ! 10^ .9h converges.
n 1 n
=
1. Look in the summation expression for a series you recognize that can be used for your bench-
mark series. You should recognize ! .9 as a convergent geometric series, because r, namely
n
0.9, is between 0 and 1.
n 3
3
2. Use the direct comparison test to compare ! 10^ .9h to ! .9 . First, you can pull the 10 out
n
n 1 n n 1
=
=
and ignore it because multiplying a series by a constant has no effect on its convergence or
divergence, giving you ! .9 n .
3
n 1 n
=
Now, because each term of ! .9 n is less than or equal to the corresponding term of the
3
=
n 1 n
3
3
3
convergent series ! .9 , ! .9 n has to converge as well. Finally, because ! .9 n converges,
n
=
=
n 1 n 1 n n 1 n
=
n
so does ! 10^ .9h .
3
n 1 n
=
3
h ! . 1 1 n diverges.
=
n 1 10 n
1. Find an appropriate benchmark series. Like in problem 7, there is a geometric series in the
numerator, ! . 1 1 . By the geometric series rule, it diverges. But unlike problem 7, this doesn’t
3
n
n 1
=
help you, because the given series is less than this divergent geometric series. Use the series
in the denominator instead.
3
3
3
3
! . 1 1 n = 1 ! . 1 1 n . The denominator of ! . 1 1 n is the divergent p-series ! .
1
n 1 10 n 10 n 1 n n 1 n 3 n 1 n
=
=
=
=
2. Apply the direct comparison test. Because each term of ! . 1 1 n is greater than the
n 1 n
=
3
3
corresponding term of the divergent series ! , ! . 1 1 n diverges as well — and therefore
1
n 1 n n 1 n
=
=
3
so does ! . 1 1 n .
n 1 10 n
=
1
1
1
1
i 1001 + 2001 + 3001 + 4001 + ... diverges.
1 1 1 1
1. Ask yourself what this series resembles: It’s the divergent harmonic series: + + + + ....
1 2 3 4
2. Multiply the given series by 1001 so that you can compare it to the harmonic series.
1 1 1 1 1001 1001 1001 1001
1001c + + + + ... = + + + + ...
m
1001 2001 3001 4001 1001 2001 3001 4001