in this section, we consider positive series

The next three sections develop techniques for determining whether an infinite series converges or diverges. This is easier than finding the sum of an infinite series, which is possible only in special cases.

In this section, we consider positive serieswhere an > 0 for all n. We can visualize the terms of a positive series as rectangles of width 1 and height an (Figure 1).The partial sumis equal to the area of the first N rectangles.

The key feature of positive series is that their partial sums form an increasing sequence:

for all N. This is because SN+1 is obtained from SN by adding a positive number:

1N NS S

1 2N nS a a a

1 1 2 1 1,N N N N NS a a a a S a 1 0Na

Recall that an increasing sequence converges if it is bounded above. Otherwise, it diverges (Theorem 6, Section 10.1). It follows that a positive series behaves in one of two ways (this is the dichotomy referred to in the next theorem).

THEOREM 1 Dichotomy for Positive Series

1

is a positive seriesnn

S a

(i) The partial sums SN are bounded above. In this case, S converges. Or,(ii) The partial sums SN are not bounded above. In this case, S diverges.

• Theorem 1 remains true if an ≥ 0. It is not necessary to assume that an > 0.

• It also remains true if an > 0 for all n ≥ M for some M, because the convergence of a series is not affected by the first M terms.

Assumptions Matter The dichotomy does not hold for a nonpositive series. Consider

The partial sums are bounded (because SN = 1 or 0), but S diverges.

1

1

1 1 1 1 1 1 1n

n

S

Our first application of Theorem 1 is the following Integral Test. It is extremely useful because integrals are easier to evaluate than series in most cases.THEOREM 2 Integral Test Let an = f (n), where f (x) is positive, decreasing, and continuous for x ≥ 1.

11

11

i converges converges.

ii diverges diverges.

nn

nn

f x dx a

f x dx a

The Harmonic Series Diverges Show that diverges.

an = f (n), where f (x) is positive, decreasing, and continuous for x ≥ 1.

1Let :f xx

1 1

lim lim ln ,R

R R

dx dx Rx x

1

1 diverges.n n

We will use the fact that the harmonic series diverges, quite often.

THEOREM 2 Integral Test Let an = f (n), where f (x) is positive, decreasing, and continuous for x ≥ 1.

11

11

i converges converges.

ii diverges diverges.

nn

nn

f x dx a

f x dx a

2

2 22 221 1 2

1 1 1lim lim lim2 21 1

RR R

R R R

x xdx dx u duux x

2 2 2 221

1 2 3Does converge?2 5 101n

n

n

2 1 2u x du xdx

2

1 1 1lim lim2 2

14 4

R

R Ru R

221

converges.1n

n

n

The sum of the reciprocal powers n−p is called a p-series.

THEOREM 3 Convergence of p-Series The infinite seriesconverges if p > 1 and diverges otherwise.

Here are two examples of p-series:

Another powerful method for determining convergence of positive series is comparison. Suppose that 0 ≤ an ≤ bn. Figure 4 suggests that if the larger sum converges, then the smaller sum also converges. Similarly, if the smaller sum diverges, then the larger sum also diverges.

THEOREM 4 Comparison Test Assume that there exists M > 0 such that 0 ≤ an ≤ bn for n ≤ M.

1 1

1 1

i converges converges.

ii diverges diverges.

n nn n

n nn n

b a

a b

1

1Does converge?3nn n

1 1 1, 33 nn

nn

THEOREM 4 Comparison Test Assume that there exists M > 0 such that 0 ≤ an ≤ bn for n ≤ M.

1 1

1 1

i converges converges.

ii diverges diverges.

n nn n

n nn n

b a

a b

1 1

1 1 converges converges3 3n n

n n n

THEOREM 2 Sum of a Geometric Series Let c 0. If |r| < 1, then 2 3

0 1n

n

ccr c cr cr crr

1/322

1Does converge?3n

Sn

1/32

1/32 3 2

3 2

2

1 1 , 23

3 3 0

We must show 3 0, 2

2' 3 2 3 2 0 0,3

2 is increasing 3

nn n

n n n n

f x x x x

f x x x x x x

f x

2 1 1, 2f f x x

The divergence of(called the harmonic series) was known to the medieval scholar Nicole d’Oresme (1323–1382).

1/322

the larger series 1 also diverges.3n n

Using the Comparison Correctly Study the convergence of

2

We cannot compare the series to the harmonic series,1 1because the harmonic series diverges & .

lnn n n

Fortunately, the Integral Test can be used. The substitution u = ln x yields

22

1The integral test shows that converges.lnn n n

2 1x dx Cx

Suppose we wish to study the convergence of

Thus we might try to compare S with

Unfortunately, however, the inequality goes in the wrong direction:

Although the smaller series

converges, we cannot use the Comparison Theorem to say anything about our larger series. In this situation, the following variation of the Comparison Test can be used.

THEOREM 5 Limit Comparison Test Let {an} and {bn} be positive sequences. Assume that the following limit exists:

** 0 converges converges

** and converges converges

** 0 and converges converges

n n

n n

n n

L a b

L a b

L b a

CONCEPTUAL INSIGHT To remember the different cases of the Limit Comparison Test, you can think of it this way. If L > 0, then an ≈ Lbn for large n. In other words, the series

and n na b are roughly multiples of each other, so one converges if and only if the other converges. If L = ∞, then an is much larger than bn (for large n), so if

converges, certainly coverges.n na b Finally, if L = 0, then bn is much larger an and the convergence of

yields the convergence of .n nb a

2

42

Show that converges.1n

nn n

THEOREM 5 Limit Comparison Test Let {an} and {bn} be positive sequences. Assume that the following limit exists:

** 0 converges converges

** and converges converges

** 0 and converges converges

n n

n n

n n

L a b

L a b

L b a

lim 1 ** 0 converges convergesnn nn

n

a L a bb

lim n

nn

aLb

2

2 42 2

1Since converges, also converges by THM 5.1n n

nn n n

23

1Determine whether coverges.4n n

THEOREM 5 Limit Comparison Test Let {an} and {bn} be positive sequences. Assume that the following limit exists:

** 0 converges converges

** and converges converges

** 0 and converges converges

n n

n n

n n

L a b

L a b

L b a

lim n

nn

aLb

lim 1 ** 0 converges convergesnn nn

n

a L a bb

22 2

1 1Since diverges, also diverges by THM 5.4n nn n