In 7-12, compute the limits using the technique shown in Example 5.
7.
10.
8.
11.
9.
12.
13. Suppose a power series
converges for
for all other values of . What is the value of and why?
14. Suppose that a power series
series converge faster for
15. Consider
converges for
. Will the power
? Explain your reasoning.
or for
.
a. Compute the radius of convergence for
.
b. Differentiate term by term to find the series for
convergence.
c. Let
be the function such that
series expansion for
(around
16. Let
and diverges
, and then find its radius of
and
. Find the power
), and find its radius of convergence.
.
a. The series for
converges like a geometric series. What is the ratio for this
geometric series?
b. Using
terms from the sum for
we think we have an error of less than
. How many terms should we sum to get an error less than
?
c. Is your estimate on the number of terms in part (b) likely to be too large or too
small? Explain your reasoning.
17. Let
.
a. The series for
converges like a geometric series. What is the ratio for this
geometric series?
b. Using
terms from the sum for
we think we have an error of less than
. How many terms should we sum to get an error less than
?
c. Is your estimate on the number of terms in part (b) likely to be too large or too
small? Explain your reasoning.
9
These rules both make sense in terms of the growth rates for different powers of , and can
be justified with algebra. We give one example to illustrate how an algebraic proof would
work.
Example 5:
The main idea is to multiply both top and bottom by the same quantity, in this case
Then, the limit of each part of the expression is easy to compute individually.
.
The rules listed above can all be justified in general with this technique. In applying the
ratio test, feel free to use the rules listed above without further explanation.
Finally, we have described how one may determine limits of rational functions of
when the numerator and denominators have leading coefficients of 1. While other cases
are unlikely to arise from the ratio test, they are not any more difficult to compute. For
example,
Exercises
In 1-6, determine the interval of convergence for each power series. You do not have to
determine whether or not the series converge at the endpoints.
1.
4.
2.
5.
3.
6.
8
This means that for
, the terms in the sum
have ratios approaching
. Thus, the convergence rate should be roughly the same as
for a geometric series with ratio
. For comparison, here is an analogous table for
such a geometric series:
Geometric series:
10
20
30
40
50
60
Sum of terms
1.3333320617675781250
1.3333333333321206737
1.3333333333333333322
1.3333333333333333333
1.3333333333333333333
1.3333333333333333333
Error
Although the errors themselves are not the same, the rate at which the errors are going
to 0 is the same. For the geometric series, each error is exactly
times as large
as the previous error (we look at
because increases by 10 from one line of the tables
to the next). For the power series, the errors should go down by approximately the same
factor, and they do.
3
Tips for taking limits
The limits which arise from the ratio test often contain rational functions of . In other
words, one has to compute limits of the sort
lower power terms
lower power terms
Since we are interested in the behavior of the fraction when
is large, we suspect that the
lower power terms do not affect the outcome and that the result is the same as for
This is exactly what happens
when the numerator and denominator have the same degree (
.
), the limit is 1;
when the degree of the numerator, , is less than the degree of the denominator, , the
limit is 0;
when the degree of the numerator, , is greater than the degree of the denominator, ,
the limit is
.
7
We are able to reduce the error to a simple formula because it is a geometric series too!
Note, the error does go to 0 as gets large because
(and and are fixed).
As gets large, this expression goes to 0 exponentially. There are a number of ways
of interpreting this statement. For example, if we use terms, and then add more terms
and compare the errors:
Error
Error
So using an addition
terms in the sum reduces the error by a factor of
(which is less
than since
).
Another way of interpreting this is that number of correct decimal places when adding
terms in the sum is roughly proportional to the number of terms added. In particular, for
large if we double we will double the number of correct decimal places in the sum.
Now suppose we have a power series and a value of for which the series converges by
the ratio test (i.e., not an endpoint value). If we sum the first terms of the power series,
we cannot compute the error exactly, so we estimate it by analogy with a geometric series.
The same conclusion as above applies : the number of correct digits from adding terms
of the series is roughly proportional to as gets large. The constant of proportionality
will depend on the function and the value of . Moreover, adding
terms will multiply
the error by approximately
where is the value of the limit computed in the ratio test..
We can observe these phenomenon numerically. Consider the series from Example 4,
Taylor series for
sums for
at
terms for
:
. The table shows the values of the
along with the errors from these approximations.
Power series with
10
20
30
40
50
60
Sum of terms
-0.28768204431685190352
-0.28768207245176670366
-0.28768207245178092743
-0.28768207245178092744
-0.28768207245178092744
-0.28768207245178092744
Error
There are several things worth observing in the table. First, the errors go to 0 very
quickly. This makes the power series a useful tool for computing
. Second, we
notice that the magnitude of the error (the exponent on the power of 10) is decreasing at a
steady rate. This corresponds to the statement that the number of correct decimal places is
proportional to the number of terms used for the approximation.
Finally, we could have predicted the rate at which the error went to 0. When we applied
the ratio test to this series, we found that the ratios of successive terms approached
.
To be honest, there is one case for which this discussion is not quite right. When the radius of convergence
is infinite, the series converges even faster than this: the power series converges faster than any geometric
series.
6
2.2 Radius of convergence when integrating and differentiating power
series
As discussed in 10.3 of the text, one can integrate or differentiate one power series to get
another. The radius of convergence remains the same under either operation. It is not hard
to demonstrate this fact algebraically with the ratio test. Here we look at a typical example.
For the series
, the ratio test tells us that the series converges when
So, the radius of convergence is . If we differentiate this series term by term we get the
new series
and compute its radius of convergence with the ratio test:
The result looks very similar. The only difference in the limits is the extra factor of
which goes to 1 as goes to infinity. Thus the radius of convergence is the same as it was for
the original series. This approach can be applied to either the integration or differentiation
of any power series.
2.3 Rates of convergence
One of the applications of power series is to compute values of functions. It is only necessary to sum a finite number of terms from a series as an approximation to the exact value.
To make good use of this technique, we have to understand the behavior of the error when
only a finite number of terms from a series are used. The basic idea is that convergent
power series behave like geometric series (except at the endpoints of their intervals of convergence). We can analyze the behavior of geometric series completely, and then transfer
the information to a general power series.
In all discussion of convergent power series in this section, we will assume that we are
not dealing with an endpoint value.
Convergence of a geometric series can be effectively analyzed since we have a simple
form for its sum:
if
The error from estimating this sum by using only the first
Error
Exact Value
Sum of first
5
terms
terms is:
From the ratio test, we can then conclude that the original series converges when
,
in other words, when
. This is an interval centered at 1 of radius 1. The series
must diverge when
(i.e., when
or
). For this series, our method
does not tell us what happens at the endpoints – when
or when
.
Remark: As we see from this example, the power series need not be centered at 0 when
applying the ratio test.
2
Theoretical conclusions based on the ratio test
There are a number of phenomena related to convergence of power series which can be
explained in terms of the ratio test. In this section, we assume that we are dealing with a
power series as in the examples above .
2.1 Existence of radius of convergence
As you have seen in the text, the set of values where a power series converges is always an
interval centered at , the central point for the power series. The distance from to the
edge of this interval is called the radius of convergence.
The fact that the domain of a power series is always an interval centered at is forced
by the type of inequality produced by the ratio test. That is, if we apply the ratio test to
:
Setting this limit less than , there are cases:
typically,
is a positive real number. Then we find that the series will
converge when
where
and the test is not conclusive on the endpoints
the power series converges is an interval of length
. It diverges when
,
. In any case, the set where
centered at .
if
, then the power series converges for all . This is what happened in Example 1. We think of the radius of convergence being infinite in this case.
There are no endpoints to consider.
if
, then the series converges only when
which makes
every term in the sum identically . We would say that the radius of convergence
here is and there are no endpoints to consider. This is extremely unusual, especially
when dealing with Taylor series. We leave it to the reader to check that it can occur
for the series
.
It turns out that the series converges at
but that it diverges at
.
With a few technical improvements, the arguments presented here really work for any power series.
4
From the ratio test, we can then conclude that the original series converges when
(i.e., when
). The series must diverge when
(i.e., when
). The test is not conclusive for the values
or
.
or
Example 3:
Again we apply the ratio test and compute:
By the ratio test, the series will converge when
, i.e., when
.
It will diverge when
or when
. The values
are not yet
determined .
This computation required us to evaluate two limits. The limit of
and similar
limits are discussed in 3 of this handout. Similar reasoning can be applied to the other
limit
as well, or note that
The final equality comes by evaluating the limit for each term of the expression individually.
Example 4:
This sum arises as the Taylor series of
at
. Applying the ratio test, we find
It is not hard to see that the series diverges in each of these cases by other means. For example, when
we are looking at the sum
. Since the terms of this sum do not tend to 0,
the sum will diverge. The case of
is similar.
The series converges at both
. The method for showing this is through what is called the integral
test, a comparison between the sum and an improper integral.
3
This sum arises as the Taylor series of around
sum converges, we apply the ratio test with
. To determine for which values the
and compute the limit:
In the last line, the limit is because is fixed and is going to infinity. Since the limit is
always less than , the series converges for all .
Remark: As noted above, this power series is the Taylor series for at
. While the
ratio test assures us that the sum
converges for all values of , it does not guarantee
that the sum converges to . This is accomplished in 10.5 of the text by analyzing the
error formula for Taylor series. When dealing with a Taylor series of a function
:
the error term for Taylor series can be used to establish that a Taylor series converges
to
on an interval. An automatic consequence is that the Taylor series does
converge on this interval.
the ratio test will determine where a Taylor series converges to something. It is
possible that it does not converge to the original function
, but in practice this
is relatively rare. Moreover, the ratio test is much easier to use than Taylor' s error
formula.
Example 2:
We compute the limit prescribed by the ratio test.
Remember that in taking this limit,
is the only variable which is changing, and we find
which is equal to 1 (see 3).
2
The Ratio Test
A power series is an expression of the form
(1)
where
and the are constants. Power series arise in a number of contexts; the Taylor
series of a function is the most prominent in calculus. A power series such as (1) defines
a function for all values of where this sum converges. We would like to determine these
values of (i.e., the domain of the power series). The ratio test provides a quick method of
determining almost all of these values. Here is the statement:
The Ratio Test Given a series
where
for
if
, then the sum
converges;
if
, then the sum
diverges.
sufficiently large,
Remark: if
, the theorem makes no claims at all; the test is not conclusive in this case and one would have to use other means to determine whether or not the
sum converges.
Remark: If
, then the ratio test applies implying that the series
diverges.
The idea behind the ratio test is that if
, then for large each
. In other words, the series will behave like a geometric series with ratio . A geometric
series converges iff its ratio satisfies
.
1
Using the ratio test
We illustrate how the ratio test can be used to determine the domain of a power series.
As we will see, the values of for which a power series converges is always an interval.
We will be concerned with determining this interval with the ratio test, but not with the
question of whether or not the power series converges at the endpoints of this interval .
Example 1:
For the interested reader, some remarks on endpoints are contained in footnotes throughout.
1