James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

744 Chapter 11 Infinite Sequences and Series
47. (a) Find the partial sum s 5 of the series o ǹ−1 1ysn2n d. Use Exercise
46 to estimate the error in using s 5 as an approximation
to the sum of the series.
(b) Find a value of n so that s n is within 0.00005 of the sum.
Use this value of n to approximate the sum of the series.
48. Use the sum of the first 10 terms to approximate the sum of
the series
n
2 n
ò
n−1
Use Exercise 46 to estimate the error.
49. Prove the Root Test. [Hint for part (i): Take any number r such
that L , r , 1 and use the fact that there is an integer N such
, r whenever n > N.]
that s n | an |
50. Around 1910, the Indian mathematician Srinivasa Ramanujan
discovered the formula
1
− 2s2
9801 ò s4nd!s1103 1 26390nd
n−0 sn!d 4 396 4n
William Gosper used this series in 1985 to compute the first
17 million digits of .
(a) Verify that the series is convergent.
(b) How many correct decimal places of do you get if you
use just the first term of the series? What if you use two
terms?
51. Given any series o a n, we define a series o a n
1
whose terms are
all the positive terms of o a n and a series o a n
2 whose terms
are all the negative terms of o a n. To be specific, we let
a n
1 − an 1 | an |
2
a n
2 − an 2 | an |
2
Notice that if a n . 0, then a n1 − a n and a n
2 − 0, whereas if
a n , 0, then a n
2− a n and a n 1 − 0.
(a) If o a n is absolutely convergent, show that both of the
series o a n
1
and o a n
2 are convergent.
(b) If o a n is conditionally convergent, show that both of the
series o a n
1
and o a n
2 are divergent.
52. Prove that if o a n is a conditionally convergent series and
r is any real number, then there is a rearrangement of o a n
whose sum is r. [Hints: Use the notation of Exercise 51.
Take just enough positive terms a n
1 so that their sum is greater
than r. Then add just enough negative terms a n
2 so that the
cumulative sum is less than r. Continue in this manner and use
Theorem 11.2.6.]
53. Suppose the series o a n is conditionally convergent.
(a) Prove that the series o n 2 a n is divergent.
(b) Conditional convergence of o a n is not enough to determine
whether o na n is convergent. Show this by giving an
example of a conditionally convergent series such that
o na n converges and an example where o na n diverges.
We now have several ways of testing a series for convergence or divergence; the problem
is to decide which test to use on which series. In this respect, testing series is similar
to integrating functions. Again there are no hard and fast rules about which test to apply
to a given series, but you may find the following advice of some use.
It is not wise to apply a list of the tests in a specific order until one finally works. That
would be a waste of time and effort. Instead, as with integration, the main strategy is to
classify the series according to its form.
1. If the series is of the form o1yn p , it is a p-series, which we know to be convergent
if p . 1 and divergent if p < 1.
2. If the series has the form o ar n21 or o ar n , it is a geometric series, which converges
if | r | , 1 and diverges if | r | > 1. Some preliminary algebraic manipulation
may be required to bring the series into this form.
3. If the series has a form that is similar to a p-series or a geometric series, then
one of the comparison tests should be considered. In particular, if a n is a rational
function or an algebraic function of n (involving roots of polynomials), then the
series should be compared with a p-series. Notice that most of the series
in Exercises 11.4 have this form. (The value of p should be chosen as in Section
11.4 by keeping only the highest powers of n in the numerator and denominator.)
The comparison tests apply only to series with positive terms, but if o a n
has some negative terms, then we can apply the Comparison Test to o| a n | and
test for absolute convergence.
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