APEX Ratio and Root Tests

Section 9.4 Ratio and Root Tests

The

n

th-Term Test of Theorem 9.2.23 states that in order for a series

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

to converge,

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

That is, the terms of

{a_{n}}

must get very small. Not only must the terms approach 0, they must approach 0 “fast enough”: while

lim_{n \to \infty} 1 / n = 0,

the Harmonic Series

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

diverges as the terms of

{1 / n}

do not approach 0 “fast enough.”

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The comparison tests of the previous section determine convergence by comparing terms of a series to terms of another series whose convergence is known. This section introduces the Ratio and Root Tests, which determine convergence by analyzing the terms of a series to see if they approach 0 “fast enough.”

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Subsection 9.4.1 Ratio Test

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Theorem 9.4.1. Ratio Test.

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Let

{a_{n}}

be a positive sequence and consider

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

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If $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} < 1,$ then $\sum_{n = 1}^{\infty} a_{n}$ converges.
If $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} > 1$ or $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = \infty,$ then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
If $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1,$ the Ratio Test is inconclusive.

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Figure 9.4.2. Video presentation of Theorem 9.4.1

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The principle of the Ratio Test is this: if

lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = L < 1,

then for large

n,

each term of

{a_{n}}

is significantly smaller than its previous term which is enough to ensure convergence.

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Example 9.4.3. Applying the Ratio Test.

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Use the Ratio Test to determine the convergence of the following series:

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

Solution 1.

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

$\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{2^{n + 1} / (n + 1)!}{2^{n} / n!} \\ = lim_{n \to \infty} \frac{2^{n + 1} n!}{2^{n} (n + 1)!} \\ = lim_{n \to \infty} \frac{2}{n + 1} \\ = 0 . \end{aligned}$

Since the limit is $0 < 1,$ by the Ratio Test $\sum_{n = 1}^{\infty} \frac{2^{n}}{n!}$ converges. The fact that $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 0$ can be interpreted to mean that in the long run, the term $a_{n + 1}$ is roughly $0$ times as large as $a_{n} .$ In other words, not only is $a_{n}$ decreasing to $0,$ it is decreasing very quickly. That is, the terms of $a_{n}$ decrease to $0$ sufficiently fast enough to guarantee the convergence of $\sum_{n = 1}^{\infty} a_{n} .$
$\sum_{n = 1}^{\infty} \frac{3^{n}}{n^{3}} :$

$\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{3^{n + 1} / (n + 1)^{3}}{3^{n} / n^{3}} \\ = lim_{n \to \infty} \frac{3^{n + 1} n^{3}}{3^{n} (n + 1)^{3}} \\ = lim_{n \to \infty} \frac{3 n^{3}}{(n + 1)^{3}} \\ = 3 . \end{aligned}$

Since the limit is $3 > 1,$ by the Ratio Test $\sum_{n = 1}^{\infty} \frac{3^{n}}{n^{3}}$ diverges. The fact that $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 3$ can be interpreted to mean that in the long run, the term $a_{n + 1}$ is roughly $3$ times as large as $a_{n},$ so $a_{n}$ is increasing by roughly a factor of $3$ in the long run. We could also use Theorem 9.2.23 to determine that this series diverges. The exponential will dominate the polynomial in the long run, so $lim_{n \to \infty} 3^{n} / n^{3} = \infty .$
$\sum_{n = 1}^{\infty} \frac{1}{n^{2} + 1} :$

$\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{1 / ((n + 1)^{2} + 1)}{1 / (n^{2} + 1)} \\ = lim_{n \to \infty} \frac{n^{2} + 1}{(n + 1)^{2} + 1} \\ = 1 . \end{aligned}$

Since the limit is 1, the Ratio Test is inconclusive. We can easily show this series converges using the Integral Test. We can also use Direct Comparison Test or Limit Comparison Test, with each comparing to the series $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} .$

Solution 2. Video solution

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The Ratio Test is not effective when the terms of a series only contain algebraic functions (e.g., polynomials). It is most effective when the terms contain some factorials or exponentials. The previous example also reinforces our developing intuition: factorials dominate exponentials, which dominate algebraic functions, which dominate logarithmic functions. In Part 1 of the example, the factorial in the denominator dominated the exponential in the numerator, causing the series to converge. In Part 2, the exponential in the numerator dominated the algebraic function in the denominator, causing the series to diverge.

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While we have used factorials in previous sections, we have not explored them closely and one is likely to not yet have a strong intuitive sense for how they behave. The following example gives more practice with factorials.

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Example 9.4.4. Applying the Ratio Test.

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Determine the convergence of

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

Solution 1.

Before we begin, be sure to note the difference between

(2 n)!

and

2 n! .

When

n = 4,

the former is

8! = 8 \cdot 7 \cdot \dots \cdot 2 \cdot 1 = 40, 320,

whereas the latter is

2 (4 \cdot 3 \cdot 2 \cdot 1) = 48 .

Applying the Ratio Test:

\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{(n + 1)! (n + 1)! / (2 (n + 1))!}{n! n! / (2 n)!} \\ = lim_{n \to \infty} \frac{(n + 1)! (n + 1)! (2 n)!}{n! n! (2 n + 2)!} \end{aligned}

Noting that

(n + 1)! = (n + 1) \cdot n!

and

(2 n + 2)! = (2 n + 2) \cdot (2 n + 1) \cdot (2 n)!,

we have

\begin{aligned} = lim_{n \to \infty} \frac{(n + 1) (n + 1)}{(2 n + 2) (2 n + 1)} \\ = 1 / 4 . \end{aligned}

Since the limit is

1 / 4 < 1,

by the Ratio Test we conclude

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

converges.

To find the limit in the second to last line, recall that we just need to examine the leading terms of the numerator and denominator, which are

n^{2}

and

4 n^{2}

respectively.

Solution 2. Video solution

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Subsection 9.4.2 Root Test

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The final test we introduce is the Root Test, which works particularly well on series where each term is raised to a power, and does not work well with terms containing factorials.

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Theorem 9.4.5. Root Test.

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Let

{a_{n}}

be a positive sequence, and consider

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

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If $lim_{n \to \infty} (a_{n})^{1 / n} < 1,$ then $\sum_{n = 1}^{\infty} a_{n}$ converges.
If $lim_{n \to \infty} (a_{n})^{1 / n} > 1$ or $lim_{n \to \infty} (a_{n})^{1 / n} = \infty,$ then $\sum_{n = 1}^{\infty} a_{n}$ diverges.
If $lim_{n \to \infty} (a_{n})^{1 / n} = 1,$ the Root Test is inconclusive.

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Figure 9.4.6. Video presentation of Theorem 9.4.5

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Example 9.4.7. Applying the Root Test.

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Determine the convergence of the following series using the Root Test:

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$\sum_{n = 1}^{\infty} {(\frac{3 n + 1}{5 n - 2})}^{n}$
$\sum_{n = 1}^{\infty} \frac{n^{4}}{(\ln (n))^{n}}$
$\sum_{n = 1}^{\infty} \frac{2^{n}}{n^{2}}$

Solution 1.

$\begin{aligned} lim_{n \to \infty} {(a_{n})}^{1 / n} & = lim_{n \to \infty} {({(\frac{3 n + 1}{5 n - 2})}^{n})}^{1 / n} \\ = lim_{n \to \infty} \frac{3 n + 1}{5 n - 2} = \frac{3}{5} . \end{aligned}$

Since the limit is less than 1, we conclude the series converges. Note: it is difficult to apply the Ratio Test to this series.
$\begin{aligned} lim_{n \to \infty} {(a_{n})}^{1 / n} & = lim_{n \to \infty} {(\frac{n^{4}}{(\ln (n))^{n}})}^{1 / n} \\ = lim_{n \to \infty} \frac{(n^{4 / n})}{\ln (n)} \end{aligned}$
The limit of the numerator must be found using L’Hospital’s Rule for indeterminate powers
$\begin{aligned} lim_{n \to \infty} (n^{4 / n}) & = lim_{n \to \infty} e^{\ln (n^{4 / n})} \\ = lim_{n \to \infty} e^{4 \ln (n) / n} \end{aligned}$
Now apply L’Hospital’s to the expression in the exponent:
$\begin{aligned} \overset{by LHR}{=} lim_{n \to \infty} e^{4 / n} \\ = e^{0} = 1 . \end{aligned}$

Since the numerator approaches 1 (by L’Hospital’s Rule) and the denominator grows to infinity, we have

$lim_{n \to \infty} \frac{(n^{4 / n})}{\ln (n)} = 0 .$

Since the limit is less than 1, we conclude the series converges.
$lim_{n \to \infty} {(\frac{2^{n}}{n^{2}})}^{1 / n} = lim_{n \to \infty} \frac{2}{(n^{2 / n})} = 2 .$ Since this is greater than 1, we conclude the series diverges. (Note: The Ratio Test is easy to apply to this series.)

(Also note: The limit in the denominator is found in a similar fashion as was illustrated in Part 2. In general $lim_{n \to \infty} (n)^{b / n} = 1$ for any real number $b .$ )

Solution 2. Video solution

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Each of the tests we have encountered so far has required that we analyze series from positive sequences. Section 9.5 relaxes this restriction by considering alternating series, where the underlying sequence has terms that alternate between being positive and negative.

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The Ratio Test is not effective when the terms of a sequence only contain functions.

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2.

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The Ratio Test is most effective when the terms of a sequence contains and/or functions.

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3.

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What three convergence tests do not work well with terms containing factorials?

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4.

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The Root Test works particularly well on series where each term is to a .

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Problems

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Exercise Group.

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In the following exercises, determine the convergence of the given series using the Ratio Test. If the Ratio Test is inconclusive, state so and determine convergence with another test.

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5.

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\sum_{n = 0}^{\infty} \frac{2 n}{n!}

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6.

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\sum_{n = 0}^{\infty} \frac{5^{n} - 3 n}{4^{n}}

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7.

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\sum_{n = 0}^{\infty} \frac{n! 10^{n}}{(2 n)!}

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8.

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\sum_{n = 1}^{\infty} \frac{5^{n} + n^{4}}{7^{n} + n^{2}}

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9.

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

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10.

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\sum_{n = 1}^{\infty} \frac{1}{3 n^{3} + 7}

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11.

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\sum_{n = 1}^{\infty} \frac{10 \cdot 5^{n}}{7^{n} - 3}

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12.

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\sum_{n = 1}^{\infty} n \cdot {(\frac{3}{5})}^{n}

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13.

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\sum_{n = 1}^{\infty} \frac{2 \cdot 4 \cdot 6 \cdot 8 \dots 2 n}{3 \cdot 6 \cdot 9 \cdot 12 \dots 3 n}

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14.

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\sum_{n = 1}^{\infty} \frac{n!}{5 \cdot 10 \cdot 15 \dots (5 n)}

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Exercise Group.

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In the following exercises, determine the convergence of the given series using the Root Test. If the Root Test is inconclusive, state so and determine convergence with another test.

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