Section: PT系列:对趋同的比重和根测试 | 9.8 PT系列:对聚合的比重和根试验

Section outline

When the $\begin{aligned} n \end{aligned}$ th partial sum of an infinite series approaches a finite limit as $\begin{aligned} n \end{aligned}$ goes to infinity, the series is said to converge . Convergence may occur in a positive term series when each successive term of the series is less than the previous term, i.e. when the ratio of terms (current to previous) is less than 1. Testing for this condition is the basis for the Ratio Test. If a ratio can be determined to be less than 1 we can have convergence; a ratio greater than 1 means terms are growing and the series will diverge . Do you think this kind of test will work for a positive term series like $\begin{aligned} \sum_{n = 1}^{\infty} \frac{3}{n} \end{aligned}$ ?
::当无限序列的 n 部分总和接近n 到无穷的限定值时,据说该系列会趋同。当该系列的每个连续任期少于前一任期时,即当条件比率(目前与以前的比率)小于1时,会在一个正数系列中出现趋同。这个条件的测试是该比率测试的基础。如果能确定一个比率小于1,我们就能达到趋同;一个大于1的数值正在增加,而该系列则会不同。你认为这种试验对正数系列,如n=13n会有效吗?

Ratio and Root Tests for Convergence
::聚合比率和根测试

So far we have looked at the following tests for the convergence/divergence of infinite series:
::迄今为止,我们研究了下列无限系列趋同/共振的测试:

Convergence Test	Applicable Series
Limit of $\begin{aligned} n \end{aligned}$ th partial sum	All
$\begin{aligned} n \end{aligned}$ th-term test for divergence (and $\begin{aligned} n \end{aligned}$ th-term property for convergent series)	All
	Positive term
Comparison Tests: ::比较测试: Direct Comparison Test ::直接比较测试 Limit Comparison Test (and Simplified Limit Comparison Test) ::限制性比较试验(和简化的限制性比较试验)	Positive term

In this concept we take a look at two more tests that are used for positive term series:
::在这个概念中,我们再研究一下用于正用语系列的另外两项检验:

The Ratio Test
::比率测试

The Root (or $\begin{aligned} n \end{aligned}$ th Root) Test
::根( 或nth根) 测试

The Ratio Test
::比率测试

The Ratio Test looks at the ratio of terms of the series to make a determination of convergence or divergence. This test may be useful when the terms or the series have factorials or powers of $\begin{aligned} n \end{aligned}$ .
::比率测试查看系列条款的比例,以确定趋同或差异。当术语或系列具有n的系数或权力时,这一测试可能有用。

The Ratio Test is as follows:
::比率测试如下:

Let $\begin{aligned} \sum_{n = 1}^{\infty} a_{n} \end{aligned}$ be a positive term series (i.e., $\begin{aligned} a_{n} > 0 \end{aligned}$ ) .
::让我们让“n=1”成为正术语序列(即 an>0) 。

If $\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = L < 1 \end{aligned}$ , then the series is convergent.
::如果limnan+1an=L<1,则序列为集合。

If $\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = L > 1 \end{aligned}$ , then the series is divergent.
::如果 limnan+1an=L>1, 那么序列是不同的。

If $\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = 1 \end{aligned}$ , then the series may be convergent or divergent; another test is needed.
::如果 limnan+1an=1, 则该序列可以是会合或相异的; 需要另外测试。

Note that this test may be helpful when $\begin{aligned} a_{n} \end{aligned}$ has factorials or powers of $\begin{aligned} n \end{aligned}$ .
::请注意,当某一因素或权力为n时,本检验标准可能有所帮助。

Let's use the ratio test to determine if the series $\begin{aligned} \sum \frac{3^{n}}{n!} \end{aligned}$ converge or diverge.
::让我们使用比率测试来确定序列 3 nn! 集合还是偏离。

The series $\begin{aligned} \sum \frac{3^{n}}{n!} \end{aligned}$ is a positive term series with $\begin{aligned} a_{n} = \frac{3^{n}}{n!} \end{aligned}$ .
::系列 =3nn! 是一个正数术语序列使用 an=3nn!

Using the Ratio Test we have,
::使用我们的比数测试,

\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{\frac{3^{n + 1}}{(n + 1)!}}{\frac{3^{n}}{n!}} \\ = lim_{n \to \infty} \frac{3^{n + 1}}{3^{n}} \frac{n!}{(n + 1)!} \\ = lim_{n \to \infty} \frac{3}{n + 1} \\ = 0 \end{aligned}

::=======================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

By the Ratio Test, the series $\begin{aligned} \sum \frac{3^{n}}{n!} \end{aligned}$ is convergent.
::以比数测试为基准,序列 ====================================================================================================================================================================================================================Y===============================================================================================================================================================================================================================================================================================

The Ratio Test is not always conclusive, and the following problems illustrate why there is a need to recognize series types and learn alternative methods for determining convergence.
::比率测试并不总是有结论性的,以下问题说明为什么需要承认系列类型,并学习其他方法确定趋同。

Let's try to use Ratio Test to determine whether each of the following series converge:
::让我们尝试使用比率测试来确定以下序列中的每一个序列是否趋同 :

The harmonic series $\begin{aligned} \sum \frac{1}{n} \end{aligned}$
::口音序列 #% 1n

The $\begin{aligned} p \end{aligned}$ -series $\begin{aligned} \sum \frac{1}{n^{2}} \end{aligned}$
::P系列 1n2

The harmonic series $\begin{aligned} \sum \frac{1}{n} \end{aligned}$ is a positive term series. Applying the Ratio Test yields
::调和序列 #% 1n 是正数术语序列。应用比率测试收益率

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

::limnan+1an=limn1(n+1)1(n+1)1n=limnnn+1=1

According to the Ratio Test, convergence/divergence is inconclusive and another test is required. We know from previous discussions that the harmonic series is divergent.
::根据比率测试,趋同/振荡是没有结果的,还需要进行另一次测试,我们从以前的讨论中知道,调和序列是不同的。

The $\begin{aligned} p \end{aligned}$ -series is a positive term series, and here $\begin{aligned} p = 2 \end{aligned}$ . Applying the Ratio Test yields
::p系列是一个正数术语序列, 这里p=2. 应用比率测试的收益率

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

::limnan+1an=limn1(n+1)21n2=limn(nn+1)2=1

According to the Ratio Test, convergence/divergence is inconclusive and another test is required. But, we know from previous discussions that for $\begin{aligned} p = 2 > 1 \end{aligned}$ , the $\begin{aligned} p \end{aligned}$ -series is convergent.
::根据比率测试,趋同/振荡是没有结果的,需要再做一次测试。但是,我们从以前的讨论中知道,p=2>1的p系列是趋同的。

The Root (or $\begin{aligned} n \end{aligned}$ th-Root) Test
::根( 或 nth- Rooot) 测试

The looks at the limit of a root of the $\begin{aligned} n \end{aligned}$ th term of the series, and may be useful when the terms have powers of $\begin{aligned} n \end{aligned}$ .
::审视该系列第 n 个任期根的限度,当术语具有n 权限时可能有用。

The Root Test is as follows:
::根测试如下:

Let $\begin{aligned} \sum_{n = 1}^{\infty} a_{n} \end{aligned}$ be a positive term series.
::让我们成为积极的术语系列。

If $\begin{aligned} lim_{n \to \infty} \sqrt[n]{a_{n}} = L < 1 \end{aligned}$ , then the series is convergent.
::如果limnann=L<1,则序列为集合。

If $\begin{aligned} lim_{n \to \infty} \sqrt[n]{a_{n}} = L > 1 \end{aligned}$ or $\begin{aligned} lim_{n \to \infty} \sqrt[n]{a_{n}} = \infty \end{aligned}$ , then the series is divergent.
::如果L1或LINANN,

If $\begin{aligned} lim_{n \to \infty} \sqrt[n]{a_{n}} = L = 1 \end{aligned}$ , then the series may be convergent or divergent; another test is needed.
::如果limnann=L=1,则该序列可以是会合或相异的;需要另外测试。

Note that this test may be helpful when $\begin{aligned} a_{n} \end{aligned}$ has powers of $\begin{aligned} n \end{aligned}$ such. For example, the Root Test is faster than the Ratio Test in determining if the series $\begin{aligned} \sum {(\frac{n + 1}{2 n + 3})}^{n} \end{aligned}$ is convergent.
::请注意,当 n 具有这种功率时,本测试可能有用。例如,根测试比比率测试更快,以确定 (n+12n+3)n 序列是否趋同。

Let $\begin{aligned} a_{n} = {(\frac{n + 1}{2 n + 3})}^{n} \end{aligned}$ .
::Let=(n+12n+3)n. Let=(n+12n+3)n.

The Root Test yields,
::根测试产量,

$\begin{aligned} lim_{n \to \infty} \sqrt[n]{| a_{n} |} = lim_{n \to \infty} \frac{n + 1}{2 n + 3} = \frac{1}{2} \end{aligned}$ .
::===================================================================================================================================================================================================================================== ====================================================================================

By the Root Test, the series $\begin{aligned} \sum {(\frac{n + 1}{2 n + 3})}^{n} \end{aligned}$ is convergent.
::根测试时, *(n+12n+3)n 序列是聚合的。

Now, using the Ratio Test for the same series, we get:
::现在,使用同一系列的比率测试,我们得到:

\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} & = lim_{n \to \infty} \frac{{(\frac{n + 2}{2 n + 5})}^{n + 1}}{{(\frac{n + 1}{2 n + 3})}^{n}} \\ = lim_{n \to \infty} (\frac{n + 2}{2 n + 5}) \cdot {[\frac{(n + 2) (2 n + 3)}{(n + 1) (2 n + 5)}]}^{n} \\ = lim_{n \to \infty} (\frac{1 + \frac{2}{n}}{2 + \frac{5}{n}}) \cdot {[\frac{(1 + \frac{2}{n}) (2 + \frac{3}{n})}{(1 + \frac{1}{n}) (2 + \frac{5}{n})}]}^{n} \\ = lim_{n \to \infty} (\frac{1 + \frac{2}{n}}{2 + \frac{5}{n}}) \cdot lim_{n \to \infty} {[\frac{(1 + \frac{2}{n}) (2 + \frac{3}{n})}{(1 + \frac{1}{n}) (2 + \frac{5}{n})}]}^{n} \\ = (\frac{1}{2}) \cdot {[\frac{(1) (2)}{(1) (2)}]}^{n} \\ = \frac{1}{2} \end{aligned}

::====================================================================================================================================================================~=================================================================================================================================================================================================================================

By the Ratio Test, the series $\begin{aligned} \sum {(\frac{n + 1}{2 n + 3})}^{n} \end{aligned}$ is convergent.
::通过比率测试, =(n+12n+3)n 序列是集合的。

The Root Test was less work! So we should learn to apply the right test.
::根测试不太有效,所以我们应该学会应用正确的测试

Examples
::实例

Example 1
::例1

Earlier, you were asked if a the Ratio Test would work for a positive term series like $\begin{aligned} \sum_{n = 1}^{\infty} \frac{3}{n} \end{aligned}$ .
::早些时候,有人问过你比率测试是否对正数序列有效比如"n=13n"

The series $\begin{aligned} \sum_{n = 1}^{\infty} \frac{3}{n} \end{aligned}$ has decreasing terms, so that $\begin{aligned} \frac{a_{n + 1}}{a_{n}} = \frac{n}{n + 1} < 1 \end{aligned}$ . Yet, this series can be shown to diverge, contrary to expectations. What’s going on? Turns out, $\begin{aligned} lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = lim_{n \to \infty} \frac{n}{n + 1} = 1 \end{aligned}$ , so that the series behavior is too close to call using this test. Another approach is necessary!
::“n”=13n系列有递减条件,所以一个+1an=nn+1<1。然而,这一系列可以显示差异,这与预期相反。到底发生了什么?结果显示, limnan+1an=limnnn+1=1, 使得该系列行为太接近于使用此测试。另一种方法是必要的 !

Example 2
::例2

Determine if the series $\begin{aligned} \sum_{n = 1}^{\infty} e^{- 2 n} n! \end{aligned}$ converges using both the Ratio Test and the Root Test.
::确定序列 'n=1\\\\e-2nn! 是否使用比率测试和根测试组合。

The series is a positive term series with $\begin{aligned} a_{n} = e^{- 2 n} n! \end{aligned}$ .
::这个系列是一个正面的术语序列, 使用 an=e- 2nn!

Applying the Ratio Test yields,
::应用比率测试产量,

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

::=================================================================================================+=========================================================================================================================================================================================

According to the Ratio Test, the series $\begin{aligned} \sum_{n = 1}^{\infty} e^{- 2 n} n! \end{aligned}$ diverges .
::根据比率测试,序列=1=1e-2nn!

Applying the Root Test yields,
::应用根测试产量,

\begin{aligned} lim_{n \to \infty} \sqrt[n]{a_{n}} & = lim_{n \to \infty} \sqrt[n]{e^{- 2 n} n!} \\ = lim_{n \to \infty} e^{- 2} \sqrt[n]{n!} \\ = e^{- 2} lim_{n \to \infty} \sqrt[n]{n!} \end{aligned}

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!

This evaluation is more difficult than for the Ratio Test because finding $\begin{aligned} lim_{n \to \infty} \sqrt[n]{n!} \end{aligned}$ is not easy.
::此项评估比比率测试更困难, 因为找到limnn! n 并不容易。

It can be shown (e.g., by using Stirling’s approximation: $\begin{aligned} n! \approx \sqrt{2 π n} {(\frac{n}{e})}^{n} \end{aligned}$ ) that $\begin{aligned} lim_{n \to \infty} \sqrt[n]{n!} = \infty \end{aligned}$ .
::使用 Stirling 的近似值 : n! @\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

The Root Test and Ratio Test give consistent results; but the Ratio Test was so much easier to use.
::根测试和比率测试得出一致的结果;但比率测试使用起来容易得多。

Review
::回顾

For #1-8, determine whether the series converges or diverges using the Ratio Test.
::对于 # 1-8, 使用比率测试来确定该序列是否趋同或不同。

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n}{5^{n}} \end{aligned}$
::=1 n5n

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n^{2}}{2^{n}} \end{aligned}$
::=1 n22n

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n!}{e^{n}} \end{aligned}$
::来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来,来!

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n!}{1 \cdot 3 \cdot 5 \dots (2 n - 1)} \end{aligned}$
::-==YTET -伊甸园字幕组=-伊甸园字幕组=- 翻译:

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n}{3^{n}} \end{aligned}$
::=1 n3n

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n!}{n^{n}} \end{aligned}$
::不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不,不!

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{n - 1}{n 4^{n}} \end{aligned}$
::=1 -1n4n =1n -1n4n =1n -1n4n

For #9-15, determine whether the series converges or diverges using the Root Test.
::对于# 9-15, 确定序列是否使用根测试相趋近或不同。

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{e^{n}}{n^{n}} \end{aligned}$
::=1 nnn

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{2^{n - 3}}{4^{n}} \end{aligned}$
::=1 2n-34n

$\begin{aligned} \sum_{n = 1}^{\infty} {(\frac{n^{2} + 1}{2 n^{2} + 1})}^{n} \end{aligned}$
::n=1(n2+12n2+1)n

$\begin{aligned} \sum_{n = 1}^{\infty} \frac{3^{2 n}}{7^{n}} \end{aligned}$
::=1 =3千2n7n

$\begin{aligned} \sum_{n = 1}^{\infty} {(\frac{\ln (n^{2} + 1)}{n})}^{n} \end{aligned}$
::n=1( ln( n2+1)n) n

$\begin{aligned} \sum_{n = 1}^{\infty} {(\frac{1}{n} - \frac{1}{n^{2}})}^{n} \end{aligned}$
::n=1(1n-1n2n)n

$\begin{aligned} \sum_{n = 1}^{\infty} n {(\frac{2}{3})}^{n} \end{aligned}$
::Nn=1n( 23)n

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

Ratio and Root Tests for Convergence ::聚合比率和根测试

The Ratio Test ::比率测试

The Root (or n th-Root) Test ::根( 或 nth- Rooot) 测试

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)