Ratio and Root Tests

In this section, we prove the last two series convergence tests: the ratio test and the root test. These tests are particularly nice because they do not require us to find a comparable series. The ratio test will be especially useful in the discussion of power series in the next chapter.

Throughout this chapter, we have seen that no single convergence test works for all series. Therefore, at the end of this section we discuss a strategy for choosing which convergence test to use for a given series.

Ratio Test

From our earlier discussion and examples, we know that

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

is not a sufficient condition for the series to converge. Not only do we need

a_{n} \to 0,

Here we introduce the ratio test, which provides a way of measuring how fast the terms of a series approach zero.

Proof

We begin with the proof of part i. In this case,

ρ = lim_{n \to \infty} \| \frac{a_{n + 1}}{a_{n}} \| < 1 .

converges. Given the inequalities above, we can apply the comparison test and conclude that the series

For part iii. we show that the test does not provide any information if

ρ = 1

The ratio test is particularly useful for series whose terms contain factorials or exponentials, where the ratio of terms simplifies the expression. The ratio test is convenient because it does not require us to find a comparative series. The drawback is that the test sometimes does not provide any information regarding convergence.

Root Test

The approach of the root test is similar to that of the ratio test. Consider a series

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

The expression on the right-hand side is a geometric series. As in the ratio test, the series

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

the test does not provide any information. For example, for any p-series,

\sum_{n = 1}^{\infty} 1 / n^{p},

To evaluate this limit, we use the natural logarithm function. Doing so, we see that

The root test is useful for series whose terms involve exponentials. In particular, for a series whose terms

a_{n}

Choosing a Convergence Test

At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.

Using Convergence Tests

For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.

$\sum_{n = 1}^{\infty} \frac{n^{2} + 2 n}{n^{3} + 3 n^{2} + 1}$
$\sum_{n = 1}^{\infty} \frac{{(−1)}^{n + 1} (3 n + 1)}{n!}$
$\sum_{n = 1}^{\infty} \frac{e^{n}}{n^{3}}$
$\sum_{n = 1}^{\infty} \frac{3^{n}}{{(n + 1)}^{n}}$

Step 1. The series is not a $p - series$
or geometric series.

Step 2. The series is not alternating.

Step 3. For large values of
$n,$
we approximate the series by the expression

$\frac{n^{2} + 2 n}{n^{3} + 3 n^{2} + 1} \approx \frac{n^{2}}{n^{3}} = \frac{1}{n} .$

Therefore, it seems reasonable to apply the comparison test or limit comparison test using the series
$\sum_{n = 1}^{\infty} 1 / n .$
Using the limit comparison test, we see that

$lim_{n \to \infty} \frac{(n^{2} + 2 n) / (n^{3} + 3 n^{2} + 1)}{1 / n} = lim_{n \to \infty} \frac{n^{3} + 2 n^{2}}{n^{3} + 3 n^{2} + 1} = 1 .$

Since the series
$\sum_{n = 1}^{\infty} 1 / n$
diverges, this series diverges as well.
Step 1.The series is not a familiar series.

Step 2. The series is alternating. Since we are interested in absolute convergence, consider the series

$\sum_{n = 1}^{\infty} \frac{3 n}{(n + 1)!} .$

Step 3. The series is not similar to a p-series or geometric series.

Step 4. Since each term contains a factorial, apply the ratio test. We see that

$lim_{n \to \infty} \frac{(3 (n + 1)) / (n + 1)!}{(3 n + 1) / n!} = lim_{n \to \infty} \frac{3 n + 3}{(n + 1)!} \cdot \frac{n!}{3 n + 1} = lim_{n \to \infty} \frac{3 n + 3}{(n + 1) (3 n + 1)} = 0 .$

Therefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.
Step 1. The series is not a familiar series.

Step 2. It is not an alternating series.

Step 3. There is no obvious series with which to compare this series.

Step 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.

Step 5. To apply the divergence test, we calculate that

$lim_{n \to \infty} \frac{e^{n}}{n^{3}} = \infty .$

Therefore, by the divergence test, the series diverges.
Step 1. This series is not a familiar series.

Step 2. It is not an alternating series.

Step 3. There is no obvious series with which to compare this series.

Step 4. Since each term is a power of
$n,$
we can apply the root test. Since

$lim_{n \to \infty} \sqrt[n]{{(\frac{3}{n + 1})}^{n}} = lim_{n \to \infty} \frac{3}{n + 1} = 0,$

by the root test, we conclude that the series converges.

In [link], we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series

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

has negative terms, these tests can be applied to

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

Summary of Convergence Tests
Series or Test	Conclusions	Comments
Divergence Test For any series $\sum_{n = 1}^{\infty} a_{n},$ evaluate $lim_{n \to \infty} a_{n} .$	If $lim_{n \to \infty} a_{n} = 0,$ the test is inconclusive.	This test cannot prove convergence of a series.
If $lim_{n \to \infty} a_{n} \neq 0,$ the series diverges.
Geometric Series $\sum_{n = 1}^{\infty} a r^{n - 1}$	If $\\| r \\| < 1,$ the series converges to $a / (1 - r) .$	Any geometric series can be reindexed to be written in the form $a + a r + a r^{2} + ⋯,$ where $a$ is the initial term and $r$ is the ratio.
If $\\| r \\| \geq 1,$ the series diverges.
p-Series $\sum_{n = 1}^{\infty} \frac{1}{n^{p}}$	If $p > 1,$ the series converges.	For $p = 1,$ we have the harmonic series $\sum_{n = 1}^{\infty} 1 / n .$
If $p \leq 1,$ the series diverges.
Comparison Test For $\sum_{n = 1}^{\infty} a_{n}$ with nonnegative terms, compare with a known series $\sum_{n = 1}^{\infty} b_{n} .$	If $a_{n} \leq b_{n}$ for all $n \geq N$ and $\sum_{n = 1}^{\infty} b_{n}$ converges, then $\sum_{n = 1}^{\infty} a_{n}$ converges.	Typically used for a series similar to a geometric or $p$ -series. It can sometimes be difficult to find an appropriate series.
If $a_{n} \geq b_{n}$ for all $n \geq N$ and $\sum_{n = 1}^{\infty} b_{n}$ diverges, then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
Limit Comparison Test For $\sum_{n = 1}^{\infty} a_{n}$ with positive terms, compare with a series $\sum_{n = 1}^{\infty} b_{n}$ by evaluating $L = lim_{n \to \infty} \frac{a_{n}}{b_{n}} .$	If $L$ is a real number and $L \neq 0,$ then $\sum_{n = 1}^{\infty} a_{n}$ and $\sum_{n = 1}^{\infty} b_{n}$ both converge or both diverge.	Typically used for a series similar to a geometric or $p$ -series. Often easier to apply than the comparison test.
If $L = 0$ and $\sum_{n = 1}^{\infty} b_{n}$ converges, then $\sum_{n = 1}^{\infty} a_{n}$ converges.
If $L = \infty$ and $\sum_{n = 1}^{\infty} b_{n}$ diverges, then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
Integral Test If there exists a positive, continuous, decreasing function $f$ such that $a_{n} = f (n)$ for all $n \geq N,$ evaluate $\int_{N}^{\infty} f (x) d x .$	$\int_{N}^{\infty} f (x) d x$ and $\sum_{n = 1}^{\infty} a_{n}$ both converge or both diverge.	Limited to those series for which the corresponding function $f$ can be easily integrated.
Alternating Series $\sum_{n = 1}^{\infty} {(−1)}^{n + 1} b_{n} or \sum_{n = 1}^{\infty} {(−1)}^{n} b_{n}$	If $b_{n + 1} \leq b_{n}$ for all $n \geq 1$ and $b_{n} \to 0,$ then the series converges.	Only applies to alternating series.
Ratio Test For any series $\sum_{n = 1}^{\infty} a_{n}$ with nonzero terms, let $ρ = lim_{n \to \infty} \\| \frac{a_{n + 1}}{a_{n}} \\| .$	If $0 \leq ρ < 1,$ the series converges absolutely.	Often used for series involving factorials or exponentials.
If $ρ > 1 or ρ = \infty,$ the series diverges.
If $ρ = 1,$ the test is inconclusive.
Root Test For any series $\sum_{n = 1}^{\infty} a_{n},$ let $ρ = lim_{n \to \infty} \sqrt[n]{\\| a_{n} \\|} .$	If $0 \leq ρ < 1,$ the series converges absolutely.	Often used for series where $\\| a_{n} \\| = b_{n}^{n} .$
If $ρ > 1 or ρ = \infty,$ the series diverges.
If $ρ = 1,$ the test is inconclusive.

Series Converging to

π

and

1 / π

Dozens of series exist that converge to $π$

or an algebraic expression containing $π .$

Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of $π$

in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.

The series

$π = 4 \sum_{n = 1}^{\infty} \frac{{(−1)}^{n + 1}}{2 n - 1} = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \frac{4}{9} - ⋯$

was discovered by Gregory and Leibniz in the late
1600s.
This result follows from the Maclaurin series for
f(x)=tan−1x.
We will discuss this series in the next chapter.
1. Prove that this series converges.
2. Evaluate the partial sums $S_{n}$
  for
  $n = 10, 20, 50, 100 .$
3. Use the remainder estimate for alternating series to get a bound on the error $R_{n} .$
4. What is the smallest value of $N$
  that guarantees
  $\| R_{N} \| < 0.01 ?$
  Evaluate
  $S_{N} .$
The series

$\begin{matrix} π & = 6 \sum_{n = 0}^{\infty} \frac{(2 n)!}{2^{4 n + 1} {(n!)}^{2} (2 n + 1)} \\ = 6 (\frac{1}{2} + \frac{1}{2 \cdot 3} {(\frac{1}{2})}^{3} + \frac{1 \cdot 3}{2 \cdot 4 \cdot 5} \cdot {(\frac{1}{2})}^{5} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 7} {(\frac{1}{2})}^{7} + ⋯) \end{matrix}$

has been attributed to Newton in the late
1600s.
The proof of this result uses the Maclaurin series for
f(x)=sin−1x.
1. Prove that the series converges.
2. Evaluate the partial sums $S_{n}$
  for
  $n = 5, 10, 20 .$
3. Compare $S_{n}$
  to
  $π$
  for
  $n = 5, 10, 20$
  and discuss the number of correct decimal places.
The series

$\frac{1}{π} = \frac{\sqrt{8}}{9801} \sum_{n = 0}^{\infty} \frac{(4 n)! (1103 + 26390 n)}{{(n!)}^{4} 396^{4 n}}$

was discovered by Ramanujan in the early
1900s.
William Gosper, Jr., used this series to calculate
π
to an accuracy of more than
17
million digits in the
mid-1980s.
At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for
π
and
1/π.
1. Prove that this series converges.
2. Evaluate the first term in this series. Compare this number with the value of $π$
  from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?
3. Investigate the life of Srinivasa Ramanujan $(1887 – 1920)$
  and write a brief summary. Ramanujan is one of the most fascinating stories in the history of mathematics. He was basically self-taught, with no formal training in mathematics, yet he contributed in highly original ways to many advanced areas of mathematics.

Key Concepts

the test does not provide any information. This test is useful for series whose terms involve factorials.

the test does not provide any information. The root test is useful for series whose terms involve powers.

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$

converges, where $a_{n}$

is given in the following problems. State if the ratio test is inconclusive.

a_{n} = 1 / n!

a_{n + 1} / a_{n} \to 0 .

Converges.

a_{n} = 10^{n} / n!

a_{n} = n^{2} / 2^{n}

\frac{a_{n + 1}}{a_{n}} = \frac{1}{2} {(\frac{n + 1}{n})}^{2} \to 1 / 2 < 1 .

Converges.

a_{n} = n^{10} / 2^{n}

\sum_{n = 1}^{\infty} \frac{{(n!)}^{3}}{(3 n!)}

\frac{a_{n + 1}}{a_{n}} \to 1 / 27 < 1 .

Converges.

\sum_{n = 1}^{\infty} \frac{2^{3 n} {(n!)}^{3}}{(3 n!)}

\sum_{n = 1}^{\infty} \frac{(2 n)!}{n^{2 n}}

\frac{a_{n + 1}}{a_{n}} \to 4 / e^{2} < 1 .

Converges.

\sum_{n = 1}^{\infty} \frac{(2 n)!}{{(2 n)}^{n}}

\sum_{n = 1}^{\infty} \frac{n!}{{(n / e)}^{n}}

\frac{a_{n + 1}}{a_{n}} \to 1 .

Ratio test is inconclusive.

\sum_{n = 1}^{\infty} \frac{(2 n)!}{{(n / e)}^{2 n}}

\sum_{n = 1}^{\infty} \frac{{(2^{n} n!)}^{2}}{{(2 n)}^{2 n}}

\frac{a_{n}}{a_{n + 1}} \to 1 / e^{2} .

Converges.

Use the root test to determine whether $\sum_{n = 1}^{\infty} a_{n}$

converges, where $a_{n}$

is as follows.

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

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

{(a_{k})}^{1 / k} \to 2 > 1 .

Diverges.

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

a_{n} = n / 2^{n}

{(a_{n})}^{1 / n} \to 1 / 2 < 1 .

Converges.

a_{n} = n / e^{n}

a_{k} = \frac{k^{e}}{e^{k}}

{(a_{k})}^{1 / k} \to 1 / e < 1 .

Converges.

a_{k} = \frac{π^{k}}{k^{π}}

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

a_{n}^{1 / n} = \frac{1}{e} + \frac{1}{n} \to \frac{1}{e} < 1 .

Converges.

a_{k} = \frac{1}{{(1 + ln k)}^{k}}

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

a_{n}^{1 / n} = \frac{(ln (1 + ln n))}{(ln n)} \to 0

by L’Hôpital’s rule. Converges.

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series $\sum_{k = 1}^{\infty} a_{k}$

with given terms $a_{k}$

converges, or state if the test is inconclusive.

a_{k} = \frac{k!}{1 \cdot 3 \cdot 5 ⋯ (2 k - 1)}

a_{k} = \frac{2 \cdot 4 \cdot 6 ⋯ 2 k}{(2 k)!}

\frac{a_{k + 1}}{a_{k}} = \frac{1}{2 k + 1} \to 0 .

Converges by ratio test.

a_{k} = \frac{1 \cdot 4 \cdot 7 ⋯ (3 k - 2)}{3^{k} k!}

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

{(a_{n})}^{1 / n} \to 1 / e .

Converges by root test.

a_{k} = {(\frac{1}{k + 1} + \frac{1}{k + 2} + ⋯ + \frac{1}{2 k})}^{k}

(Hint: Compare $a_{k}^{1 / k}$

to $\int_{k}^{2 k} \frac{d t}{t} .)$

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

a_{k}^{1 / k} \to ln (3) > 1 .

Diverges by root test.

a_{n} = {(n^{1 / n} - 1)}^{n}

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$

converges, or state if the ratio test is inconclusive.

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

\frac{a_{n + 1}}{a_{n}} =

\frac{3^{2 n + 1}}{2^{3 n^{2} + 3 n + 1}} \to 0 .

Converge.

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

Use the root and limit comparison tests to determine whether $\sum_{n = 1}^{\infty} a_{n}$

converges.

a_{n} = 1 / x_{n}^{n}

where $x_{n + 1} = \frac{1}{2} x_{n} + \frac{1}{x_{n}},$

x_{1} = 1

(Hint: Find limit of ${x_{n}} .)$

Converges by root test and limit comparison test since $x_{n} \to \sqrt{2} .$

In the following exercises, use an appropriate test to determine whether the series converges.

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

\sum_{n = 1}^{\infty} \frac{{(−1)}^{n + 1} (n + 1)}{n^{3} + 3 n^{2} + 3 n + 1}

Converges absolutely by limit comparison with $p - series,$

p = 2 .

\sum_{n = 1}^{\infty} \frac{{(n + 1)}^{2}}{n^{3} + {(1.1)}^{n}}

\sum_{n = 1}^{\infty} \frac{{(n - 1)}^{n}}{{(n + 1)}^{n}}

lim_{n \to \infty} a_{n} = 1 / e^{2} \neq 0 .

Series diverges.

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

(Hint: ${(1 + \frac{1}{n^{2}})}^{n^{2}} \approx e .)$

a_{k} = 1 / 2^{{sin}^{2} k}

Terms do not tend to zero: $a_{k} \geq 1 / 2,$

since ${sin}^{2} x \leq 1 .$

a_{k} = 2^{− sin (1 / k)}

a_{n} = 1 / (\begin{matrix} n + 2 \\ n \end{matrix})

where $(\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{k! (n - k)!}$

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

which converges by comparison with $p - series$

for $p = 2 .$

a_{k} = 1 / (\begin{array}{l} 2 k \\ k \end{array})

a_{k} = 2^{k} / (\begin{array}{l} 3 k \\ k \end{array})

a_{k} = \frac{2^{k} 1 \cdot 2 ⋯ k}{(2 k + 1) (2 k + 2) ⋯ 3 k} \leq {(2 / 3)}^{k}

converges by comparison with geometric series.

a_{k} = {(\frac{k}{k + ln k})}^{k}

(Hint: $a_{k} = {(1 + \frac{ln k}{k})}^{− (k / ln k) ln k} \approx e^{− ln k} .)$

a_{k} = {(\frac{k}{k + ln k})}^{2 k}

(Hint: $a_{k} = {(1 + \frac{ln k}{k})}^{− (k / ln k) ln k^{2}} .)$

a_{k} \approx e^{− ln k^{2}} = 1 / k^{2} .

Series converges by limit comparison with $p - series,$

p = 2 .

The following series converge by the ratio test. Use summation by parts, $\sum_{k = 1}^{n} a_{k} (b_{k + 1} - b_{k}) = [a_{n + 1} b_{n + 1} - a_{1} b_{1}] - \sum_{k = 1}^{n} b_{k + 1} (a_{k + 1} - a_{k}),$

to find the sum of the given series.

\sum_{k = 1}^{\infty} \frac{k}{2^{k}}

(Hint: Take $a_{k} = k$

and $b_{k} = 2^{1 - k} .)$

\sum_{k = 1}^{\infty} \frac{k}{c^{k}},

where $c > 1$

(Hint: Take $a_{k} = k$

and $b_{k} = c^{1 - k} / (c - 1) .)$

If $b_{k} = c^{1 - k} / (c - 1)$

and $a_{k} = k,$

then $b_{k + 1} - b_{k} = − c^{− k}$

and $\sum_{n = 1}^{\infty} \frac{k}{c^{k}} = a_{1} b_{1} + \frac{1}{c - 1} \sum_{k = 1}^{\infty} c^{− k} = \frac{c}{{(c - 1)}^{2}} .$

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

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

6 + 4 + 1 = 11

The kth term of each of the following series has a factor $x^{k} .$

Find the range of $x$

for which the ratio test implies that the series converges.

\sum_{k = 1}^{\infty} \frac{x^{k}}{k^{2}}

\sum_{k = 1}^{\infty} \frac{x^{2 k}}{k^{2}}

\| x \| \leq 1

\sum_{k = 1}^{\infty} \frac{x^{2 k}}{3^{k}}

\sum_{k = 1}^{\infty} \frac{x^{k}}{k!}

\| x \| < \infty

Does there exist a number $p$

such that $\sum_{n = 1}^{\infty} \frac{2^{n}}{n^{p}}$

converges?

Let $0 < r < 1 .$

For which real numbers $p$

does $\sum_{n = 1}^{\infty} n^{p} r^{n}$

converge?

All real numbers $p$

by the ratio test.

Suppose that $lim_{n \to \infty} \| \frac{a_{n + 1}}{a_{n}} \| = p .$

For which values of $p$

must $\sum_{n = 1}^{\infty} 2^{n} a_{n}$

converge?

Suppose that $lim_{n \to \infty} \| \frac{a_{n + 1}}{a_{n}} \| = p .$

For which values of $r > 0$

is $\sum_{n = 1}^{\infty} r^{n} a_{n}$

guaranteed to converge?

r < 1 / p

Suppose that $\| \frac{a_{n + 1}}{a_{n}} \| \leq {(n + 1)}^{p}$

for all $n = 1, 2,…$

where $p$

is a fixed real number. For which values of $p$

is $\sum_{n = 1}^{\infty} n! a_{n}$

guaranteed to converge?

For which values of $r > 0,$

if any, does $\sum_{n = 1}^{\infty} r^{\sqrt{n}}$

converge? (Hint: $\sum_{n = 1}^{\infty} a_{n} = \sum_{k = 1}^{\infty} \sum_{n = k^{2}}^{{(k + 1)}^{2} - 1} a_{n} .)$

0 < r < 1 .

Note that the ratio and root tests are inconclusive. Using the hint, there are $2 k$

terms $r^{\sqrt{n}}$

for $k^{2} \leq n < {(k + 1)}^{2},$

and for $r < 1$

each term is at least $r^{k} .$

Thus, $\sum_{n = 1}^{\infty} r^{\sqrt{n}} = \sum_{k = 1}^{\infty} \sum_{n = k^{2}}^{{(k + 1)}^{2} - 1} r^{\sqrt{n}}$

\geq \sum_{k = 1}^{\infty} 2 k r^{k},

which converges by the ratio test for $r < 1 .$

For $r \geq 1$

the series diverges by the divergence test.

Suppose that $\| \frac{a_{n + 2}}{a_{n}} \| \leq r < 1$

for all $n .$

Can you conclude that $\sum_{n = 1}^{\infty} a_{n}$

converges?

Let $a_{n} = 2^{− [n / 2]}$

where $[x]$

is the greatest integer less than or equal to $x .$

Determine whether $\sum_{n = 1}^{\infty} a_{n}$

converges and justify your answer.

One has $a_{1} = 1,$

a_{2} = a_{3} = 1 / 2,… a_{2 n} = a_{2 n + 1} = 1 / 2^{n} .

The ratio test does not apply because $a_{n + 1} / a_{n} = 1$

if $n$

is even. However, $a_{n + 2} / a_{n} = 1 / 2,$

so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if $lim_{n \to \infty} \frac{a_{2 n}}{a_{n}} < 1 / 2,$

then $\sum a_{n}$

converges, while if $lim_{n \to \infty} \frac{a_{2 n + 1}}{a_{n}} > 1 / 2,$

then $\sum a_{n}$

diverges.

Let $a_{n} = \frac{1}{4} \frac{3}{6} \frac{5}{8} ⋯ \frac{2 n - 1}{2 n + 2} = \frac{1 \cdot 3 \cdot 5 \dots (2 n - 1)}{2^{n} (n + 1)!} .$

Explain why the ratio test cannot determine convergence of $\sum_{n = 1}^{\infty} a_{n} .$

Use the fact that $1 - 1 / (4 k)$

is increasing $k$

to estimate $lim_{n \to \infty} \frac{a_{2 n}}{a_{n}} .$

Let $a_{n} = \frac{1}{1 + x} \frac{2}{2 + x} ⋯ \frac{n}{n + x} \frac{1}{n} = \frac{(n - 1)!}{(1 + x) (2 + x) ⋯ (n + x)} .$

Show that $a_{2 n} / a_{n} \leq e^{− x / 2} / 2 .$

For which $x > 0$

does the generalized ratio test imply convergence of $\sum_{n = 1}^{\infty} a_{n} ?$

(Hint: Write $2 a_{2 n} / a_{n}$

as a product of $n$

factors each smaller than $1 / (1 + x / (2 n)) .)$

a_{2 n} / a_{n} = \frac{1}{2} \cdot \frac{n + 1}{n + 1 + x} \frac{n + 2}{n + 2 + x} ⋯ \frac{2 n}{2 n + x} .

The inverse of the $k th$

factor is $(n + k + x) / (n + k) > 1 + x / (2 n)$

so the product is less than ${(1 + x / (2 n))}^{− n} \approx e^{− x / 2} .$

Thus for $x > 0,$

\frac{a_{2 n}}{a_{n}} \leq \frac{1}{2} e^{− x / 2} .

The series converges for $x > 0 .$

Let $a_{n} = \frac{n^{ln n}}{{(ln n)}^{n}} .$

Show that $\frac{a_{2 n}}{a_{n}} \to 0$

as $n \to \infty .$

Chapter Review Exercises

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit.

Is the series convergent or divergent? If convergent, is it absolutely convergent?

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula

x_{n + 1} = b x_{n},

is the average number of offspring per housefly who survive to the next generation. Assume a starting population

x_{0} .

Ratio and Root Tests

Ratio Test

Proof

Root Test

Choosing a Convergence Test

Key Concepts

Chapter Review Exercises

Glossary