The Divergence and Integral Tests

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums

{S_{k}} .

In practice, explicitly calculating this limit can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or divergence for many types of series. In this section, we discuss two of these tests: the divergence test and the integral test. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.

Divergence Test

This test is known as the divergence test because it provides a way of proving that a series diverges.

It is important to note that the converse of this theorem is not true. That is, if

lim_{n \to \infty} a_{n} = 0,

we cannot make any conclusion about the convergence of

\sum_{n = 1}^{\infty} a_{n} .

diverges. In this section and the remaining sections of this chapter, we show many more examples of such series. Consequently, although we can use the divergence test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if

a_{n} \to 0,

Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums

{S_{k}}

In this section we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the integral test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when we are considering a series whose terms are all positive.

To illustrate how the integral test works, use the harmonic series as an example. In [link], we depict the harmonic series by sketching a sequence of rectangles with areas

1, 1 / 2, 1 / 3, 1 / 4,…

We show how an integral can be used to prove that this series converges. In [link], we sketch a sequence of rectangles with areas

1, 1 / 2^{2}, 1 / 3^{2},…

is increasing and bounded, by the Monotone Convergence Theorem, it converges. Therefore, the series

\sum_{n = 1}^{\infty} 1 / n^{2}

We can extend this idea to prove convergence or divergence for many different series. Suppose

\sum_{n = 1}^{\infty} a_{n}

is an increasing sequence, if it is also a bounded sequence, then by the Monotone Convergence Theorem, it converges. We conclude that if

\int_{1}^{\infty} f (x) d x

is an unbounded sequence and therefore diverges. As a result, the series

\sum_{n = 1}^{\infty} a_{n}

it does not imply that the value of the integral and the series are the same. They may be different, and often are. For example,

The p-Series

In general, it is difficult, if not impossible, to compute the exact value of most

p

-series. However, we can use the tests presented thus far to prove whether a

p

Estimating the Value of a Series

converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum

\sum_{n = 1}^{N} a_{n}

is any positive integer. The question we address here is, for a convergent series

\sum_{n = 1}^{\infty} a_{n},

be the remainder when the sum of an infinite series is approximated by the

N th

For some types of series, we are able to use the ideas from the integral test to estimate

R_{N} .

We illustrate [link] in [link]. In particular, by representing the remainder

R_{N} = a_{N + 1} + a_{N + 2} + a_{N + 3} + ⋯

as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by

\int_{N}^{\infty} f (x) d x

Key Concepts

Key Equations

For each of the following sequences, if the divergence test applies, either state that $lim_{n \to \infty} a_{n}$

does not exist or find $lim_{n \to \infty} a_{n} .$

If the divergence test does not apply, state why.

a_{n} = \frac{n}{n + 2}

a_{n} = \frac{n}{5 n^{2} - 3}

lim_{n \to \infty} a_{n} = 0 .

Divergence test does not apply.

a_{n} = \frac{n}{\sqrt{3 n^{2} + 2 n + 1}}

a_{n} = \frac{(2 n + 1) (n - 1)}{{(n + 1)}^{2}}

lim_{n \to \infty} a_{n} = 2 .

Series diverges.

a_{n} = \frac{{(2 n + 1)}^{2 n}}{{(3 n^{2} + 1)}^{n}}

a_{n} = \frac{2^{n}}{3^{n / 2}}

lim_{n \to \infty} a_{n} = \infty

(does not exist). Series diverges.

a_{n} = \frac{2^{n} + 3^{n}}{10^{n / 2}}

a_{n} = e^{−2 / n}

lim_{n \to \infty} a_{n} = 1 .

Series diverges.

a_{n} = cos n

a_{n} = tan n

lim_{n \to \infty} a_{n}

does not exist. Series diverges.

a_{n} = \frac{1 - {cos}^{2} (1 / n)}{{sin}^{2} (2 / n)}

a_{n} = {(1 - \frac{1}{n})}^{2 n}

lim_{n \to \infty} a_{n} = 1 / e^{2} .

Series diverges.

a_{n} = \frac{ln n}{n}

a_{n} = \frac{{(ln n)}^{2}}{\sqrt{n}}

lim_{n \to \infty} a_{n} = 0 .

Divergence test does not apply.

State whether the given $p$

-series converges.

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}}

\sum_{n = 1}^{\infty} \frac{1}{n \sqrt{n}}

Series converges, $p > 1 .$

\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{2}}}

\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{4}}}

Series converges, $p = 4 / 3 > 1 .$

\sum_{n = 1}^{\infty} \frac{n^{e}}{n^{π}}

\sum_{n = 1}^{\infty} \frac{n^{π}}{n^{2 e}}

Series converges, $p = 2 e - π > 1 .$

Use the integral test to determine whether the following sums converge.

\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n + 5}}

\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n + 5}}

Series diverges by comparison with $\int_{1}^{\infty} \frac{d x}{{(x + 5)}^{1 / 3}} .$

\sum_{n = 2}^{\infty} \frac{1}{n ln n}

\sum_{n = 1}^{\infty} \frac{n}{1 + n^{2}}

Series diverges by comparison with $\int_{1}^{\infty} \frac{x}{1 + x^{2}} d x .$

\sum_{n = 1}^{\infty} \frac{e^{n}}{1 + e^{2 n}}

\sum_{n = 1}^{\infty} \frac{2 n}{1 + n^{4}}

Series converges by comparison with $\int_{1}^{\infty} \frac{2 x}{1 + x^{4}} d x .$

\sum_{n = 2}^{\infty} \frac{1}{n {ln}^{2} n}

Express the following sums as $p$

-series and determine whether each converges.

\sum_{n = 1}^{\infty} 2^{− ln n}

(Hint: $2^{− ln n} = 1 / n^{ln 2}$

2^{− ln n} = 1 / n^{ln 2} .

Since $ln 2 < 1,$

diverges by $p$

-series.

\sum_{n = 1}^{\infty} 3^{− ln n}

(Hint: $3^{− ln n} = 1 / n^{ln 3}$

\sum_{n = 1}^{\infty} n 2^{−2 ln n}

2^{−2 ln n} = 1 / n^{2 ln 2} .

Since $2 ln 2 - 1 < 1,$

diverges by $p$

-series.

\sum_{n = 1}^{\infty} n 3^{−2 ln n}

Use the estimate $R_{N} \leq \int_{N}^{\infty} f (t) d t$

to find a bound for the remainder $R_{N} = \sum_{n = 1}^{\infty} a_{n} - \sum_{n = 1}^{N} a_{n}$

where $a_{n} = f (n) .$

\sum_{n = 1}^{1000} \frac{1}{n^{2}}

R_{1000} \leq \int_{1000}^{\infty} \frac{d t}{t^{2}} = - \frac{1}{t} {\|}_{1000}^{\infty} = 0.001

\sum_{n = 1}^{1000} \frac{1}{n^{3}}

\sum_{n = 1}^{1000} \frac{1}{1 + n^{2}}

R_{1000} \leq \int_{1000}^{\infty} \frac{d t}{1 + t^{2}} = {tan}^{−1} \infty - {tan}^{−1} (1000) = π / 2 - {tan}^{−1} (1000) \approx 0.000999

\sum_{n = 1}^{100} n / 2^{n}

[T] Find the minimum value of $N$

such that the remainder estimate $\int_{N + 1}^{\infty} f < R_{N} < \int_{N}^{\infty} f$

guarantees that $\sum_{n = 1}^{N} a_{n}$

estimates $\sum_{n = 1}^{\infty} a_{n},$

accurate to within the given error.

a_{n} = \frac{1}{n^{2}},

error $< 10^{−4}$

R_{N} < \int_{N}^{\infty} \frac{d x}{x^{2}} = 1 / N, N > 10^{4}

a_{n} = \frac{1}{n^{1.1}},

error $< 10^{−4}$

a_{n} = \frac{1}{n^{1.01}},

error $< 10^{−4}$

R_{N} < \int_{N}^{\infty} \frac{d x}{x^{1.01}} = 100 N^{−0.01}, N > 10^{600}

a_{n} = \frac{1}{n {ln}^{2} n},

error $< 10^{−3}$

a_{n} = \frac{1}{1 + n^{2}},

error $< 10^{−3}$

R_{N} < \int_{N}^{\infty} \frac{d x}{1 + x^{2}} = π / 2 - {tan}^{−1} (N), N > tan (π / 2 - 10^{−3}) \approx 1000

In the following exercises, find a value of $N$

such that $R_{N}$

is smaller than the desired error. Compute the corresponding sum $\sum_{n = 1}^{N} a_{n}$

and compare it to the given estimate of the infinite series.

a_{n} = \frac{1}{n^{11}},

error $< 10^{−4},$

\sum_{n = 1}^{\infty} \frac{1}{n^{11}} = 1.000494 …

a_{n} = \frac{1}{e^{n}},

error $< 10^{−5},$

\sum_{n = 1}^{\infty} \frac{1}{e^{n}} = \frac{1}{e - 1} = 0.581976 …

R_{N} < \int_{N}^{\infty} \frac{d x}{e^{x}} = e^{− N}, N > 5 ln (10),

okay if $N = 12; \sum_{n = 1}^{12} e^{− n} = 0.581973... .$

Estimate agrees with $1 / (e - 1)$

to five decimal places.

a_{n} = \frac{1}{e^{n^{2}}},

error $< 10^{−5},$

\sum_{n = 1}^{\infty} n / e^{n 2} = 0.40488139857 …

a_{n} = 1 / n^{4},

error $< 10^{−4},$

\sum_{n = 1}^{\infty} 1 / n^{4} = π^{4} / 90 = 1.08232 ...

R_{N} < \int_{N}^{\infty} d x / x^{4} = 4 / N^{3}, N > {({4.10}^{4})}^{1 / 3},

okay if $N = 35;$

\sum_{n = 1}^{35} 1 / n^{4} = 1.08231 … .

Estimate agrees with the sum to four decimal places.

a_{n} = 1 / n^{6},

error $< 10^{−6},$

\sum_{n = 1}^{\infty} 1 / n^{4} = π^{6} / 945 = 1.01734306...,

Find the limit as $n \to \infty$

of $\frac{1}{n} + \frac{1}{n + 1} + ⋯ + \frac{1}{2 n} .$

(Hint: Compare to $\int_{n}^{2 n} \frac{1}{t} d t .)$

ln (2)

Find the limit as $n \to \infty$

of $\frac{1}{n} + \frac{1}{n + 1} + ⋯ + \frac{1}{3 n}$

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number $H_{k} = (1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{k}) .$

Recall that $T_{k} = H_{k} - ln k$

is decreasing. Compute $T = lim_{k \to \infty} T_{k}$

to four decimal places. (Hint: $\frac{1}{k + 1} < \int_{k}^{k + 1} \frac{1}{x} d x$

T = 0.5772 ...

[T] Complete sampling with replacement, sometimes called the coupon collector’s problem, is phrased as follows: Suppose you have $N$

unique items in a bin. At each step, an item is chosen at random, identified, and put back in the bin. The problem asks what is the expected number of steps $E (N)$

that it takes to draw each unique item at least once. It turns out that $E (N) = N . H_{N} = N (1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{N}) .$

Find $E (N)$

for $N = 10, 20, and 50 .$

[T] The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has $n$

cards, then the probability that the insertion will be below the card initially at the bottom (call this card $B)$

is $1 / n .$

Thus the expected number of top random insertions before $B$

is no longer at the bottom is n. Once one card is below $B,$

there are two places below $B$

and the probability that a randomly inserted card will fall below $B$

is $2 / n .$

The expected number of top random insertions before this happens is $n / 2 .$

The two cards below $B$

are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.

The expected number of random insertions to get $B$

to the top is $n + n / 2 + n / 3 + ⋯ + n / (n - 1) .$

Then one more insertion puts $B$

back in at random. Thus, the expected number of shuffles to randomize the deck is $n (1 + 1 / 2 + ⋯ + 1 / n) .$

Suppose a scooter can travel $100$

km on a full tank of fuel. Assuming that fuel can be transferred from one scooter to another but can only be carried in the tank, present a procedure that will enable one of the scooters to travel $100 H_{N}$

km, where $H_{N} = 1 + 1 / 2 + ⋯ + 1 / N .$

Show that for the remainder estimate to apply on $[N, \infty)$

it is sufficient that $f (x)$

be decreasing on $[N, \infty),$

but $f$

need not be decreasing on $[1, \infty) .$

Set $b_{n} = a_{n + N}$

and $g (t) = f (t + N)$

such that $f$

is decreasing on $[t, \infty) .$

[T] Use the remainder estimate and integration by parts to approximate $\sum_{n = 1}^{\infty} n / e^{n}$

within an error smaller than $0.0001 .$

Does $\sum_{n = 2}^{\infty} \frac{1}{n {(ln n)}^{p}}$

converge if $p$

is large enough? If so, for which $p ?$

The series converges for $p > 1$

by integral test using change of variable.

[T] Suppose a computer can sum one million terms per second of the divergent series $\sum_{n = 1}^{N} \frac{1}{n} .$

Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100 .$

[T] A fast computer can sum one million terms per second of the divergent series $\sum_{n = 2}^{N} \frac{1}{n ln n} .$

Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100 .$

N = e^{e^{100}} \approx e^{10^{43}}

terms are needed.

The Divergence and Integral Tests

Divergence Test

Integral Test

The p-Series

Estimating the Value of a Series

Key Concepts

Key Equations

Glossary