Convergence of a series containing only real positive terms

4.3 Convergence of infinite series

4.3.2 Convergence of a series containing only real positive terms

As discussed above, in order to test for the absolute convergence of a series un, we first construct the corresponding series

|un| that consists only of real positive terms. Therefore in this subsection we will restrict our attention to series of this type.

We discuss below some tests that may be used to investigate the convergence of such a series. Before doing so, however, we note the followingcrucial consideration.

In all the tests for, or discussions of, the convergence of a series, it is not what happens in the first ten, or the first thousand, or the first million terms (or any other finite number of terms) that matters, but what happensultimately.

Preliminary test

A necessarybut not sufficient condition for a series of real positive terms u_n to be convergent is that the termu_ntends to zero asn tends to infinity, i.e. we require

n→∞limu_n= 0.

If this condition is not satisfied then the series must diverge. Even if it is satisfied, however, the series may still diverge, and further testing is required.

Comparison test

The comparison test is the most basic test for convergence. Let us consider two series

u_nand

v_nand suppose that weknow the latter to be convergent (by some earlier analysis, for example). Then, if each termu_nin the first series is less than or equal to the corresponding termv_nin the second series, for alln greater than some fixed numberN that will vary from series to series, then the original series

u_nis also convergent. In other words, if

v_nis convergent and u_n≤ vn forn > N,

then

u_nconverges.

However, if

v_ndiverges andu_n≥ vnfor alln greater than some fixed number then

u_ndiverges.

Determine whether the following series converges:

∞ n=1

1 n! + 1 =1

2+1 3+1

7+ 1

25+· · · . (4.7)

Let us compare this series with the series

∞ n=0

1 n!= 1

0!+ 1 1!+ 1

2!+ 1

3!+· · · = 2 + 1 2!+ 1

3!+· · · , (4.8)

which is merely the series obtained by settingx = 1 in the Maclaurin expansion of exp x (see subsection 4.6.3), i.e.

exp(1) =e = 1 + 1 1!+ 1

2!+ 1 3!+· · · .

Clearly this second series is convergent, since it consists of only positive terms and has a finite sum. Thus, since each termunin the series (4.7) is less than the corresponding term 1/n! in (4.8), we conclude from the comparison test that (4.7) is also convergent.

D’Alembert’s ratio test

The ratio test determines whether a series converges by comparing the relative magnitude of successive terms. If we consider a series

u_nand set ρ = lim

n→∞

u_n+1 u_n



, (4.9)

then if ρ < 1 the series is convergent; if ρ > 1 the series is divergent; if ρ = 1 then the behaviour of the series is undetermined by this test.

To prove this we observe that if the limit (4.9) is less than unity, i.e.ρ < 1 then we can find a valuer in the range ρ < r < 1 and a value N such that

u_n+1 un

< r,

for alln > N. Now the terms u_nof the series that followu_N are u_N+1, u_N+2, u_N+3, . . . , and each of these is less than the corresponding term of

ru_N, r²u_N, r³u_N, . . . . (4.10) However, the terms of (4.10) are those of a geometric series with a common ratior that is less than unity. This geometric series consequently converges and therefore, by the comparison test discussed above, so must the original series u_n. An analogous argument may be used to prove the divergent case when ρ > 1.

Determine whether the following series converges:

∞ n=0

1 n!= 1

0!+ 1 1!+ 1

2!+ 1

3!+· · · = 2 + 1 2!+ 1

3!+· · · .

As mentioned in the previous example, this series may be obtained by settingx = 1 in the Maclaurin expansion of expx, and hence we know already that it converges and has the sum exp(1) =e. Nevertheless, we may use the ratio test to confirm that it converges.

Using (4.9), we have ρ = lim

n→∞

n!

(n + 1)!



= lim

n→∞

 1 n + 1



= 0 (4.11)

and sinceρ < 1, the series converges, as expected.

4.3 CONVERGENCE OF INFINITE SERIES

Ratio comparison test

As its name suggests, the ratio comparison test is a combination of the ratio and comparison tests. Let us consider the two series

u_nand

v_nand assume that we know the latter to be convergent. It may be shown that if

u_n+1 u_n ≤v_n+1

v_n for alln greater than some fixed value N then

u_nis also convergent.

Similarly, if

un+1

u_n ≥vn+1

v_n for all sufficiently largen, and

v_ndiverges then

u_nalso diverges.

Determine whether the following series converges:

∞ n=1

1

(n!)²= 1 + 1 2²+ 1

6²+· · · .

In this case the ratio of successive terms, asn tends to infinity, is given by

R = lim

n→∞

n!

(n + 1)!

2

= lim

n→∞

 1

n + 1

2

,

which is less than the ratio seen in (4.11). Hence, by the ratio comparison test, the series converges. (It is clear that this series could also be found to be convergent using the ratio test.)

Quotient test

The quotient test may also be considered as a combination of the ratio and comparison tests. Let us again consider the two series

unand

vn, and define ρ as the limit

ρ = lim

n→∞

u_n v_n



. (4.12)

Then, it can be shown that:

(i) if ρ= 0 but is finite then

u_n and 

v_n either both converge or both diverge;

(ii) ifρ = 0 and

v_nconverges then

u_nconverges;

(iii) ifρ =∞ and

v_ndiverges then

u_ndiverges.

Given that the series_∞

n=11/n diverges, determine whether the following series converges:

∞ n=1

4n²− n − 3

n³+ 2n . (4.13)

If we setun= (4n²− n − 3)/(n³+ 2n) and vn= 1/n then the limit (4.12) becomes ρ = lim

n→∞

(4n²− n − 3)/(n³+ 2n) 1/n



= lim

n→∞

4n³− n²− 3n n³+ 2n



= 4.

Sinceρ is finite but non-zero and

vndiverges, from (i) above

unmust also diverge. Integral test

The integral test is an extremely powerful means of investigating the convergence of a series

u_n. Suppose that there exists a functionf(x) which monotonically decreases forx greater than some fixed value x0and for whichf(n) = u_n, i.e. the value of the function at integer values ofx is equal to the corresponding term in the series under investigation. Then it can be shown that, if the limit of the integral

N→∞lim

 N

f(x) dx exists, the series

unis convergent. Otherwise the series diverges. Note that the integral defined here has no lower limit; the test is sometimes stated with a lower limit, equal to unity, for the integral, but this can lead to unnecessary difficulties.

Determine whether the following series converges:

∞ n=1

1

(n− 3/2)²= 4 + 4 +4 9+ 4

25+· · · .

Let us consider the functionf(x) = (x− 3/2)⁻². Clearlyf(n) = unandf(x) monotonically decreases forx > 3/2. Applying the integral test, we consider

N→∞lim

N 1

(x− 3/2)²dx = lim

N→∞

 −1

N− 3/2



= 0.

Since the limit exists the series converges. Note, however, that if we had included a lower limit, equal to unity, in the integral then we would have run into problems, since the integrand diverges atx = 3/2.

The integral test is also useful for examining the convergence of the Riemann zeta series. This is a special series that occurs regularly and is of the form

∞ n=1

1 n^p.

It converges forp > 1 and diverges if p≤ 1. These convergence criteria may be derived as follows.

4.3 CONVERGENCE OF INFINITE SERIES

Using the integral test, we consider

N→∞lim

 N 1

x^pdx = lim

N→∞

N^1−p 1− p

 ,

and it is obvious that the limit tends to zero forp > 1 and to∞ for p ≤ 1.

Cauchy’s root test

Cauchy’s root test may be useful in testing for convergence, especially if thenth terms of the series contains annth power. If we define the limit

ρ = lim

n→∞(u_n)^1/n, then it may be proved that the series

u_nconverges ifρ < 1. If ρ > 1 then the series diverges. Its behaviour is undetermined ifρ = 1.

Determine whether the following series converges:

∞ n=1

1 n

n

= 1 +1 4+ 1

27+· · · .

Using Cauchy’s root test, we find

ρ = lim

n→∞

1 n



= 0, and hence the series converges.

Grouping terms

We now consider the Riemann zeta series, mentioned above, with an alternative proof of its convergence that uses the method of grouping terms. In general there are better ways of determining convergence, but the grouping method may be used if it is not immediately obvious how to approach a problem by a better method.

First consider the case where p > 1, and group the terms in the series as follows:

SN= 1 1^p+

1 2^p+ 1

3^p

 +

1

4^p+· · · + 1 7^p

 +· · · .

Now we can see that each bracket of this series is less than each term of the geometric series

S_N= 1 1^p+ 2

2^p+ 4 4^p+· · · . This geometric series has common ratior =₁

2

p−1

; sincep > 1, it follows that r < 1 and that the geometric series converges. Then the comparison test shows that the Riemann zeta series also converges forp > 1.

The divergence of the Riemann zeta series for p ≤ 1 can be seen by first considering the casep = 1. The series is

S_N= 1 +1 2+1

3+1 4+· · · ,

which doesnot converge, as may be seen by bracketing the terms of the series in groups in the following way:

S_N=

N n=1

u_n= 1 +

1 2

 +

1 3+1

4

 +

1 5+1

6+1 7+1

8

 +· · · . The sum of the terms in each bracket is≥ ¹2and, since as many such groupings can be made as we wish, it is clear thatS_Nincreases indefinitely asN is increased.

Now returning to the case of the Riemann zeta series forp < 1, we note that each term in the series is greater than the corresponding one in the series for whichp = 1. In other words 1/n^p> 1/n for n > 1, p < 1. The comparison test then shows us that the Riemann zeta series will diverge for allp≤ 1.