## Calculus Of One Real Variable – By Pheng Kim Ving Chapter 14: Infinite Series – Section 14.7: Approximations Of Sums Of Series

14.7
Approximations Of Sums Of Series

 1. Approximations Of Sums Of Series

In previous sections of this chapter we were concerned mainly about demonstrating that a series converges or diverges. Other
than some special series like the geometric and telescoping series, we've not calculated the sum of a series after we've shown
that it converges. For most series, so far we've not developed techniques to calculate their sums. So at this stage we have to
be happy to simply settle with approximations of these sums. In this section we're going to calculate approximations of sums of
some series that converge. We'll approximate sums of some convergent series by their partial sums or by values that involve
their partial sums. We're also going to compute approximations of partial sums of the harmonic series, which diverges.

Error And Accuracy Of Approximations

When we try to approximate or estimate the sum s of a convergent series by a value A, it's natural that we want to know how
accurate the approximate or estimated value A is, that is, how close the approximate value is to the actual value. The error of
the approximation of course is the difference sA between the actual value s and the approximate value A. The accuracy is
measured by a bound on the size or magnitude (absolute value) |sA| of the error. As a matter of fact the smaller the size of
the error, the closer A is to s, and so the more accurate and thus the better the approximation. Without the knowledge of this
accuracy, the approximation wouldn't be of much use.

If we try to approximate a sum s by a value A with, for example, |error| < 0.001 (or |error| < 1/1,000), we have to find A
such that |sA| < 0.001. This of course means that A must be in the interval (s – 0.001, s + 0.001) (|sA| < 0.001 iff
– 0.001 < sA < 0.001 iff – 0.001 < As < 0.001 iff s – 0.001 < A < s + 0.001).

Number Of Decimal Digits To Use

Consider this example:

Q = 2.34567...,

A = 2.345a4a5...,

where there's no carrying of 1 from the subtraction 6 – a4 or 6 – (a4 + 1) to add to the digit 5 in A. If we use 0 or 1 or 2
decimal digits for A, (A = 2 or A = 2.3 or A = 2.34) then |E| > 0.001 (|E| = 0.34567.... or |E| = 0.04567... or |E| =
0.00567...). If we use 3 or more decimal digits (A = 2.345 or A = 2.345a4 etc), then |E| < 0.001 (|E| = 0.000d4d5...).

Now consider this example:

Q = 2.34567...,

A = 2.344a4a5...,

where there's a carrying of 1 from the subtraction 6 – a4 or 6 – (a4 + 1) to add to the 2nd digit 4 in A. If we use 0 or 1 or 2 or
3 decimal digits for A, then |E| > 0.001. If we use 4 or more decimal digits, then |E| < 0.001.

Because situations as shown in these examples are possible, we should use at least 4 significant decimal digits for A. Similarly
if Q < A, where |E| = AQ and we consider the subtraction AQ.

In general, if |error| < 1/10m, then the approximate value is accurate to at least m – 1 decimal places and we should use at
least m + 1 significant decimal digits for the approximate value.

 2. For Series Whose Convergence Is Determined By The RaT

Geometric Bound

Here we'll approximate the sums of series whose convergence is determined by the RaT (Ratio Test). We'll approximate sums
by partial sums.

Example 2.1

It can be shown that:

Notes

Solution
a. Let an = 1/n!. Then:

EOS

Remarks 2.1

d. The convergence of the series is determined by the RaT. The sum s of the series is approximated by a partial sum sn. The
bound on the error size is a geometric series. In general, for series whose convergence is determined by the RaT, it's
appropriate to approximate its sum by its partial sum and to try to bound the error size of approximation by a geometric
series. Such a bound is called a geometric bound.

 3. For Series Whose Behavior Is Determined By The IT

Here we'll approximate the sums and partial sums of positive series whose behavior (convergence or divergence) is determined
by the IT (Integral Test) or the p-series property.

Interval Containing Partial Sum Or Sum

 Fig. 3.1   Sum of areas of lower rectangles is less than or equal to area under f over [1, n].

 Fig. 3.2   Area under f over [1, n] is less than or equal to sum of areas of upper rectangles.

(Sum is between integral and integral + first term; see Fig. 3.2.)

Interval Containing Partial Sum Or Sum

 if the integral and thus the series converge.

Remark 3.1

Relation [3.3] applies to both convergent and divergent series, while relation [3.4] of course applies only to convergent series.
Relation [3.3] is useful when we desire to find an interval that has length equal to the first term of the series and that contains
the partial sum of a divergent series. Relation [3.4] is useful when we desire to find an interval that has length equal to the
first term of the series and that contains the sum of a convergent series.

Example 3.1

Solution

EOS

Example 3.2

Solution

EOS

Integral Bound

 Fig. 3.3

Approximation By sn With Integral Bound On Error

Remark 3.2

Since the error-size bound in this approximation is an integral, it's called an integral bound.

Example 3.3

Solution

So n must be at least 1,001.

EOS

Term Bound

 Fig. 3.4

Approximation With Term Bound On Error

Remarks 3.3

a. Approximation [3.6] approximates the sum by a partial sum plus another quantity, not by a partial sum alone. In this case
that other quantity is an integral,

b. The error-size bound in this approximation is a term of the series, and thus is called a term bound.

Example 3.4

Note

We'll employ the estimation:

Solution

EOS

Best Approximation With Integral Bound

From approximation [3.5] we have, for example:

 Fig. 3.5

Thus:

 Fig. 3.6   Best approximation of a quantity contained in an interval is midpoint of that interval.

Best Approximation With Integral Bound On Error Size

 The approximation of s by sn, a is called the best approximation with integral bound.

Remarks 3.4

a. Like approximation [3.6], approximation [3.8] approximates the sum by a partial sum plus another quantity, not by a partial
sum alone. In this case that other quantity is the average of 2 integrals.

b. The error bound in approximation [3.8] is an expression in terms of the integrals, and thus is called an integral bound.

c. The reason why it's called the “best approximation” will be seen below under the heading “Comparison Of The
Approximations”.

Example 3.5

Consider the best approximation of the series:

a. Using sn, determine the smallest interval that you can be sure contains its sum.
b. How large must n be so that the best approximation of the series has |error| < 0.001?
c. Find this best approximation with |error| < 0.001.

Solution

EOS

Comparison Of The Approximations

1. In Example 3.3, if s is to be approximated by sn with |error| < 0.001, then n must be at least 1,001.
1. In Example 3.4, if s is to be approximated by sn, i with |error| < 0.001, then n must be at least 32.
3. In Example 3.5, if s is to be approximated by sn, a with |error| < 0.001, then n must be at least 22.

It's easy to check that if we increase n in each approximation, the error size gets smaller. Clearly the approximation by sn, a is
the best. The approximation by sn, i comes as the relatively close second. The approximation by sn is the distant third.

Example 3.6

Solution

EOS

Relative Speed Of Convergence

Remark 3.5

Evidently the approximation of s by sn, i with the term bound on the error size or by sn or sn, a with the integral bound on the error
size is appropriate for series whose convergence is determined by the integral test or the p-series property.

 4. For Alternating Series Whose Convergence Is Determined By The AST

Here we'll approximate the sum of a convergent alternating series by its partial sums, provided the convergence of the series is
determined by the AST (alternating-series test).

 Fig. 4.1   Partial Sums Of Alternating Harmonic Series.

Approximation Of Convergent Alternating Series

 Approximating its sum by the sum of its first n terms generates an error size that is at most the magnitude of its next term.

Example 4.1

Solution
To get the accuracy of 2 decimal places we need the error size to be less than 0.005. Let s and sn be the sum and nth partial
sum of the series respectively. We have:

So at least 200 terms are needed.
EOS

This example shows that the alternating harmonic series converges very slowly.

Example 4.2

a. Show that it converges.
b. Estimate its sum to within 0.001.

Solution
a. Clearly the series is alternating, the sequence of the absolute values of its terms is decreasing, and the sequence of its
terms approaches 0 as n approaches infinity. So, by the AST, it converges.

b. We need the error size to be less than 0.001. We have:

EOS

# Problems & Solutions

1. Consider the series:

a. Show that it indeed converges.

b. Approximate its sum s by a partial sum sn so that the error size |ssn| = ssn is less than 0.0001.

Solution

a. Let an = 2n/(2n)!. Then:

So, by the RaT, the given series converges.

Note

Solution

Solution

So n must be at least 11.

4. Consider the series:

a. Show that it converges.
b. Estimate its sum with error within 0.01, using a term bound.

Solution

5. Consider the best approximation of the series:

a. Prove that this series converges.

b. Using its nth partial sum sn, determine the smallest interval that you can be sure contains its sum.
c. How large must n be so that the best approximation of the series has |error| < 0.001?
d. Find this best approximation with |error| < 0.001.

Solution

6. Consider the series:

a. Show that it converges.

b. How many terms of its terms are needed to ensure that its partial sum sn approximates its sum s with |error| < 0.001?

c. Calculate that approximate sum.

Solution

a. The series is alternating. The sequence of the absolute values of its terms is decreasing. The sequences of its terms

approaches 0. So, by the AST, it converges.

b. Its (n + 1)th term is (–1)(n+1)+1(n + 1)/2n+1 = (–1)n+2(n + 1)/2n+1. We have: