As we have observed a number of times now, we are usually unable to determine the sum of a convergent series. In fact, for most series we cannot determine whether they converge or diverge by simply looking at the sequence of partial sums. Most of the time, we will need to test a series for convergence in some indirect way. If we find that the series is convergent, we can then approximate its sum by numerically computing some partial sums. In this section, we will develop additional tests for convergence of series. The first of these is a generalization of the method we used to show that the harmonic series was divergent in Section 8.2. For a given series suppose that there is a function f  for which f (k) = ak, for k = 1, 2, ,, where f  is continuous, decreasing and f (x) 0 for all x 1. We consider the n th partial sum Look carefully at Figure 8.23a.  Figure 8.23a (n-1) rectangles, lying beneath the curve. We have constructed (n-1) rectangles on the interval [1, n], each of width 1 and with height equal to the value of the function at the right-hand endpoint of the subinterval on which the rectangle is constructed. Notice that since each rectangle lies completely beneath the curve, the sum of the areas of the (n-1) rectangles shown is less than the area under the curve from x = 1 to x = n . That is, (3.1) Note that the area of the first rectangle is length width = (1)(a2), the area of the second rectangle is (1) (a3) and so on. We get that the sum of the areas of the (n-1) rectangles indicated in Figure 8.23a is a2+ a3+ a4+ + an = Sn- a1, since Sn = a1 + a2+ + an. Together with (3.1), this gives us 0 Sum of areas of (n-1) rectangles   (3.2) Now, suppose that the improper integral converges. Then, from (3.2), we have Adding a1 to all the terms gives us This says that the sequence of partial sums is bounded. Since is also monotonic (why is that?), is convergent by Theorem 1.4 and so, the series is also convergent. In Figure 8.23b, we have constructed (n-1) rectangles on the interval [1, n], each of width 1, but with height equal to the value of the function at the left-hand endpoint of the subinterval on which the rectangle is constructed. In this case, the sum of the areas of the (n-1) rectangles shown is greater than the area under the curve. That is,   Sum of areas of (n-1) rectangles. (3.3) Figure 8.23b (n-1) rectangles, partially above the curve. Further, note that the area of the first rectangle is length width = (1) (a1), the area of the second rectangle is (1) (a2) and so on. We get that the sum of the areas of the (n-1) rectangles indicated in Figure 8.23b is  a1+a2+ +an-1 = Sn-1. Together with (3.3), this gives us 0   Sum of areas of (n-1) rectangles = Sn-1. (3.4) Now, suppose that the improper integral diverges. Since f (x) 0, this says that From (3.4), we have that This says that also. So, the sequence of partial sums diverges and hence, the series diverges, too. We summarize the results of this analysis in the following theorem.

 

3.1    Integral Test   If f (k) = ak for all k = 1, 2, and f  is continuous, decreasing and f (x) 0, for x 1, then and either both converge or both diverge.

 

It is important to recognize that while the Integral Test might say that a given series and improper integral both converge, it does not say that they will converge to the same thing. In fact, this is generally not true, as we see in the following example.

 

3.1    Using the Integral Test   Investigate the convergence or divergence of the series   The graph of the first 20 partial sums shown in Figure 8.24 suggests that the series converges to some value around 2. Figure 8.24 In the accompanying table, you can find some selected partial sums. n 10 1.97189 50 2.05648 100 2.06662 200 2.07166 500 2.07467 1000 2.07567 2000 2.07617 We have shown so many partial sums due to the very slow rate of convergence of this series. Based on our computations, we cannot say whether the series is converging very slowly to a limit around 2.076 or whether the series is instead diverging very slowly, as we had seen for the harmonic series. To determine which is the case, we must test the series further. Define Note that f  is continuous and positive everywhere and for all k 1.   Further, f ' (x) = (-1) (x2+1)-2(2x) < 0, for x (0, ), and so, f  is decreasing. This says that the Integral Test applies to this series. So, we consider the improper integral By the Integral Test, we have that since the improper integral converges, the series must converge, also. Since we have now established that the series is convergent, we can use our earlier calculations to arrive at the estimated sum 2.076. Notice that this is not the same as the value of the corresponding improper integral, which is

 

In the following example, we discuss an important type of series.

 

3.2    The p - Series   Determine for which values of p the series (a p - series) converges.     First, notice that for p = 1, this is the harmonic series, which diverges. For p > 1, define Notice that for x 1, f  is continuous and positive. Further,f ' (x) = -px-p-1 < 0, so that f  is decreasing for x 1. This says that the Integral Test applies. We now consider       Since p > 1 implies     that -p+1 < 0. In this case, the improper integral converges and so too, must the series. In the case where p < 1, we leave it as an exercise to show that the series diverges. We summarize this as follows. The p - series p - series

 

Notice that in Examples 3.1 and 3.2, we were able to use the Integral Test to establish the convergence of several series. So, now what? We have observed that you can use the partial sums of a series to estimate its value, but just how precise is a given estimate? We answer this question with the following result. First, if we estimate the sum s of the series by the nth partial sum we define the remainder Rn to be Notice that this says that the remainder is the error in approximating s by Sn. For any series shown to be convergent by the Integral Test, we can estimate the size of the remainder, as follows. From Figure 8.25, observe that the remainder Rn corresponds to the sum of the areas of the indicated rectangles.  Further, under the conditions of the Integral Test, this is less than the area under the curve y = f (x). (Recall that this area is finite, as converges.) That is, we have the following result. Figure 8.25 Estimate of the remainder

 

3.2    Error Estimate for the Integral Test   Suppose that f (k) = ak for all k = 1, 2, , where f is continuous, decreasing and f (x) 0 for all x 1. Further, suppose that converges. Then, the remainder Rn satisfies

 

We can use Theorem 3.2 to estimate the error in using a partial sum to approximate the sum of a series.

 

3.3    Estimating the Error in a Partial Sum   Estimate the error in using the partial sum S100 to approximate the sum of the series     First, recall that in Example 3.2, we had shown that this series is convergent, by the Integral Test. From Theorem 3.2 then, the remainder term satisfies

 

A more interesting and far more practical question related to Example 3.3 is to use Theorem 3.2 to help us determine the number of terms of the series necessary to obtain a given accuracy.

 

3.4    Finding the Number of Terms Needed for a Given Accuracy   Determine the number of terms needed to obtain an approximation to the sum of the series correct to within 10-5.     Again, we already used the Integral Test to show that the series in question converges. Then, by Theorem 3.2, we have that the remainder satisfies 0   So, to ensure that the remainder is less than 10-5, we require Solving this last inequality for n yields So, taking n 224 will guarantee the required accuracy and consequently, we have which is correct to within 10-5, as desired.

 

Comparison Tests We next present two results which will allow us to compare a given series with one which is already known to be convergent or divergent, much as we did with improper integrals in Section 7.7.

	3.3    Comparison Test   Suppose that ak bk, for all k. (i) If converges, then converges, too. (ii) If diverges, then diverges, too.
	Intuitively, this theorem should make abundant sense: if the “larger” series converges, then the “smaller” one must also converge. Likewise, if the “smaller” series diverges, then the “larger” one must diverge, too.

Given that ak bk for all k, observe that the n th partial sums of the two series satisfy 0 Sn = a1+a2+ +an b1+b2+ +bn. (i) If converges (say to B ), this says that (3.5) for all n 1. From (3.5), the sequence of partial sums of is bounded. Notice that is also increasing. (Why?) Since every bounded, monotonic sequence is convergent (see Theorem 1.4), we get that is convergent, too. (ii) If is divergent, we have (since all of the terms of the series are nonnegative) that Thus, must be divergent, also.

 

You can use the Comparison Test to test the convergence of series that look similar to series that you already know are convergent or divergent (notably, geometric series or p-series).

 

3.5    Using the Comparison Test for a Convergent Series   Investigate the convergence or divergence of     From the graph of the first 20 partial sums shown in Figure 8.26, it appears that the series converges to some value near 0.3. To confirm such a conjecture, we must carefully test the series. Note that for large values of k, the general term of the series looks like since when k is large, k3 is much larger than 5k.   This observation is significant, since we already know that is a convergent p - series ( p = 3 > 1 ). Further, observe that for all k 1. Since converges, the Comparison Test says that converges, too. As with the Integral Test, although the Comparison Test tells us that both series converge, the two series need not converge to the same sum. A quick calculation of a few partial sums should convince you that converges to approximately 1.202, while converges to approximately 0.2798. (Note that this is consistent with what we saw in Figure 8.26)  Figure 8.26

 

3.6    Using the Comparison Test for a Divergent Series   Investigate the convergence or divergence of   From the graph of the first 20 partial sums seen in Figure 8.27, it appears that the partial sums are growing very rapidly. On this basis, we would conjecture that the series diverges. Of course, to verify this, we need further testing. Notice that for k large, the general term looks like and we know that is a divergent geometric series Further, By the Comparison Test, diverges, too. Figure 8.27

 

There are plenty of series whose general term looks like the general term of a familiar series, but for which it is unclear how to get the inequality required for the Comparison Test to go in the right direction.

 

3.7    A Comparison That Does Not Work   Investigate the convergence or divergence of the series   Note that this is nearly identical to Example 3.5, except that there is a “ - ” sign in the denominator instead of a “ + ” sign. The graph of the first 20 partial sums seen in Figure 8.28 looks somewhat similar to the graph in Figure 8.26, except that the series appears to be converging to about 0.12. In this case, however, we have the inequality Unfortunately, this inequality goes the wrong way: we know that is a convergent p - series, but since is “larger” than this convergent series, the Comparison Test says nothing. Figure 8.28

 

Think about Example 3.7 this way: while you might observe that and you know that is convergent, the Comparison Test says nothing about the “larger” series In fact, we know that this last series is divergent (by the k th-term test for divergence, since ). To resolve this difficulty for the present problem, we will need to either make a more appropriate comparison or use the Limit Comparison Test that follows.  Notes      When we say we mean that the limit exists and is positive. In particular, we mean that

 

3.4    Limit Comparison Test   Suppose that ak, bk > 0 and that for some (finite) value, L, Then, either and both converge or both diverge.     If this says that we can make as close to L as desired. So, in particular, we can make within distance of L. That is, for some number N > 0, or (3.6) Multiplying inequality (3.6) through by bk (recall that bk > 0 ), we get Note that this says that if converges, then the “smaller” series must also converge, by the Comparison Test. Likewise, if diverges, the “larger” series must also diverge. In the same way, if converges, then converges and so, too must the “smaller” series Finally, if diverges, then diverges and hence, the “larger” series must diverge, also.

 

We can now use the Limit Comparison Test to test the series from Example 3.7 whose convergence we have so far been unable to confirm.

 

3.8    Using the Limit Comparison Test   Investigate the convergence or divergence of the series     Recall that we had already observed in Example 3.7 that the general term “looks like” for k large. We then consider the limit Since is a convergent p - series ( p = 3 > 1 ), the Limit Comparison Test says that is also convergent, as we had originally suspected.

 

The Limit Comparison Test can be used to resolve convergence questions for a great many series. The first step in using this (like the Comparison Test) is to find another series (whose convergence or divergence is known) that “looks like” the series in question.

 

3.9    Using the Limit Comparison Test   Investigate the convergence or divergence of the series The graph of the first 20 partial sums in Figure 8.29 suggests that the series converges to a limit of about 1.61. Figure 8.29 The following table of partial sums supports this conjecture. n 5 1.60522 10 1.61145 20 1.61365 50 1.61444 75 1.61453 100 1.61457 Notice that for k large, the general term looks like (since the terms with the largest exponents tend to dominate the expression, for large values of k. ) From the Limit Comparison Test, we have Since is a convergent p - series ( p = 3 > 1 ), the Limit Comparison Test says that converges, also. Finally, now that we have established thatthe series is in fact, convergent, we can use our table of computed partial sums to approximate the sum of the series as 1.61457.

 

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