Page 271 - Calculus Workbook For Dummies

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Chapter 13: Infinite Series: Welcome to the Outer Limits

Solutions to Infinite Series

3
a ! 2 n - 9 - diverges. You know (vaguely remember?) from Chapter 4 on limits that
2
n 8
2
=
n 1 5 n + 20 n + 12
2
2 n - 9 n - 8 2
lim = by the horizontal asymptote rule. Because the limit doesn’t converge
2
n " 3 5 n + 20 n + 12 5
to zero, the nth term test tells you that the series diverges.
3
b ! 1 converges to zero . . . NOT. It should be obvious that lim 1 = 0. If you conclude that the
=
n 1 n n " 3 n
3 1
series, ! , must therefore converge by the nth term test, I’ve got some bad news and some
n
n 1
=
good news for you. The bad news is that you’re wrong — you have to use the p-series test to
find out whether this converges or not (check out the solution to problem 5). The good news is
that you made this mistake here instead of on a test.
Don’t forget that the nth term test is no help in determining the convergence or divergence of a
series when the underlying sequence converges to zero.
4
c .008 - .006 + .0045 - .003375+ .00253125- ..... converges to .
875
- .006 3
1. Determine the ratio of the second term to the first term: = - .
.008 4
- 3
2. Check to see whether all the other ratios of the other pairs of consecutive terms equal .
4
.0045 3 - .003375 3 .00253125 3
- .006 = - 4 ? check . .0045 = - 4 ? check . - .003375 = - 4 ? check .
3
Voila! A geometric series with r = - .
4
3. Apply the geometric series rule.
Because 1- < r < 1, the series converges to
a = .008 = 4
1 - r 3 875
1 - - m
c
4
3
“r” is for ratio, but you may prefer, as I do, to think of r (- in this problem) as a multiplier
4
because it’s the number you multiply each term by to obtain the next.
d 1 + 1 + 1 + 1 + 1 + 1 + ... ?
2 4 8 12 16 20
1. Find the first ratio.
1
4 = 1
1 2
2
2. Test the other pairs.
1 1
8 = 1 ? check . 12 = 1 ? no .
1 2 1 2
4 8
Thus, this is not a geometric series, and the geometric series rule does not apply. Can you
guess whether this series converges or not (assuming the pattern 8, 12, 16, 20 continues)? You
can prove that this series diverges by using the limit comparison test (see problem 10) with the
harmonic series (see problem 5).