阳值和负值系列:绝对趋同比率和根测试

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Consider the alternating geometric series $\begin{align*}\sum\limits_{n=1}^{\infty} 3 \left(- \frac{2}{3} \right)^{n-1}\end{align*}$ . In the previous concept, the notion of absolute convergence was introduced by looking at the convergence of a new series created by taking the absolute value of each term , e.g., the series $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg| 3 \left(- \frac{2}{3} \right)^{n-1} \bigg|\end{align*}$ . All of the positive term convergence tests are now available to test for convergence, including the Ratio and Root Tests. What do the Ratio and Root Tests for $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg| 3 \left(- \frac{2}{3} \right)^{n-1} \bigg|\end{align*}$  tell you about its absolute convergence? What one modification of these two positive term tests would make each applicable to all series types, positive term as well as positive and negative term series?
::考虑交替几何序列 *n=13(-23n)-1。在前一个概念中,绝对趋同的概念是通过以下方式引入的:审视以每个术语的绝对值(例如,丛书) =13(-23n-1) 等每个术语的绝对值而创造的新序列的趋同。所有正值趋同测试现在都可以测试,包括比率和根值测试。n=13(-23)-3(-23n-1__) 的比值和根值测试如何告诉你其绝对趋同情况?这两个正值测试的哪些修改将使每一个都适用于所有系列、正值以及正值和正值和负值系列?

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The Ratio and Root Tests for Absolute Convergence
::绝对聚合比率和根测试

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In evaluating an infinite series containing positive and negative terms, making the series a positive term series by taking the absolute value of each term allows each of the tests for positive term series to be used to determine absolute convergence or divergence. The following two tests for the convergence of positive term series can be rewritten to reflect their use in evaluating positive and negative term series for absolute convergence:
::在评价包含正和负术语的无限系列时,通过采用每个术语的绝对值,使该系列成为正术语系列,使得每个正术语系列的测试都可用于确定绝对趋同或差异。

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• The
  ::缩略
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• The $\begin{align*}n\end{align*}$ th Root Test
  ::Nth 根测试

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The Ratio Test for Absolute Convergence
::绝对趋同比率测试

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The Ratio Test is given a slight modification when determining absolute convergence. 
::在确定绝对趋同时,对比率测试略作修改。

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The Ratio Test for Absolute Convergence  is as follows:
::绝对趋同比率测试如下:

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Let $\begin{align*}\sum\limits_{n=1}^{\infty} a_n=a_1+a_2+a_3 \ldots +a_n+ \ldots\end{align*}$  be an infinite series of non-zero numbers.
::@n=1an=a1+a2+a3...+an+...成为无限的非零数字系列。

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1. If $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n+1}{a_n} \bigg|= \alpha < 1\end{align*}$ , then the series is absolutely convergent .
   ::如果Limnan+1an1, 那么这个系列是绝对一致的 。
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2. If $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n+1}{a_n} \bigg|= \alpha > 1\end{align*}$  or $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n+1}{a_n} \bigg|= \infty\end{align*}$ , then the series is absolutely divergent.
   ::如果Limnan+1an1 或Limnan+1an1, 那么这个系列是绝对不同的。
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3. If $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n+1}{a_n} \bigg|= \alpha = 1\end{align*}$ , then the test is inconclusive.
   ::如果limnan+1an1, 那么测试是无结果的 。

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* We can ignore the zero-valued $\begin{align*}a_n\end{align*}$ ’s as far as the sum is concerned.
::* 就总金额而言,我们可以忽略零价值的一个人。

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Notice that the only difference between this version of the Ratio Test and the version introduced when working with positive term series is the use of $\begin{align*}|a_n|\end{align*}$ . When working with positive term series, there is no need to take the absolute value of the ratio of terms. However, working with series that have both positive and negative terms is different. Remember, if the use of this test shows that the series  $\begin{align*}\sum\limits_{n=1}^{\infty} |a_n|\end{align*}$ converges absolutely, then the series  $\begin{align*}\sum\limits_{n=1}^{\infty} a_n\end{align*}$ converges. If not, then another test must be used.
::请注意, 本比率测试版本和在使用正值术语序列时引入的版本的唯一区别是 {an{} 。 在使用正值术语序列时, 不需要使用该术语序列的绝对值。 但是, 与有正值和负值的序列合作是不同的。 记住, 如果使用此测试显示该序列={n=1} 和正值术语序列绝对一致, 那么序列={n=1} 就会合并。 如果没有, 那么必须使用另一个测试 。

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Always think of using the Ratio Test when a series involves a factorial such as in   $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{A^n}{n !}\end{align*}$ . Let's use the Ratio Test to determine if this series, with $\begin{align*}A\end{align*}$  a constant (positive or negative), is absolutely convergent, conditionally convergent , or divergent.
::总是会想到当序列包含一个阶乘( 如“ n= 1Ann ” 中) 时使用“ 比率测试 ! ” 。 让我们使用“ 比率测试” 来确定这个序列, 加上一个常数( 正或负) 是绝对趋同的, 有条件的趋同的, 还是不同的 。

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The Ratio Test for absolute convergence says to look at $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n+1}{a_n} \bigg|\end{align*}$ .
::绝对趋同比率测试表示要查看 limnan+1an。

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With $\begin{align*}a_n=\frac{A^n}{n !}\end{align*}$ , then
::和安!

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$$\begin{align*}\lim_{n \to \infty} \bigg|\frac{a_{n+1}}{a_n} \bigg| &=\lim_{n \to \infty} \bigg|\frac{A^{n+1}}{(n+1)!} \cdot \frac{n !}{A^n} \bigg| \\ &=\lim_{n \to \infty} \bigg|\frac{A}{n+1} \bigg| \\ &=0<1\end{align*}$$
::============================================================================================================================================== =========================================================================== ==============================================================================

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By the Ratio Test for absolute convergence, the series is absolutely convergent for any constant $\begin{align*}A\end{align*}$ , which means the series  $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{A^n}{n !}\end{align*}$ is convergent. Indeed, $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{A^n}{n !}=e^A\end{align*}$ , which is very large for large $\begin{align*}A\end{align*}$ , but still finite.
::根据绝对趋同比率测试,对于任何常数 A 来说,序列是绝对一致的, 也就是说序列 =n=1Ann! 是集合的。 事实上, =n=1Ann! =eA, 大 A 很大, 但仍有限 。

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The Root Test for Absolute Convergence
::绝对趋同的根测试

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The  $\begin{align*}n\end{align*}$ th Root Test for positive term series is given a slight modification so that it can be used to evaluate a positive and negative term series for absolute convergence.
::对正值术语序列 nth 根测试略作修改,以便用于评价正值和负值术语序列,实现绝对趋同。

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The Root Test for Absolute Convergence  is as follows:
::绝对趋同的根测试如下:

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Let $\begin{align*}\sum\limits_{n=1}^{\infty} a_n=a_1+a_2+a_3+ \ldots +a_n+ \ldots\end{align*}$  be an infinite series of non-zero numbers.
::@n=1an=a1+a2+a3+...+an+...成为无限的非零数字序列。

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1. If $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}=\alpha < 1\end{align*}$ , then the series is absolutely convergent.
   ::如果Limnnnan1, 那么这个系列是绝对一致的。
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2. If $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}=\alpha > 1\end{align*}$  or $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}=\infty\end{align*}$ , then the series is absolutely divergent.
   ::若是LIMNNNN1 或LIMNNNAN, 那么系列是绝对不同的。
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3. If $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}=1\end{align*}$ , then the test is inconclusive.
   ::如果Limnnnan1, 那么测试是无结果的 。

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Notice, again, that the only difference between this version of the $\begin{align*}n\end{align*}$ th Root Test and the version introduced when working with positive term series is the use of $\begin{align*}|a_n|\end{align*}$ . When working with positive term series, there is no need to take the absolute value of the series term. However, working with series that have both positive and negative terms is different.
::请注意,本版本的Nth根测试和在使用正值术语序列时采用的版本的唯一区别是使用 an。在使用正值术语序列时,不需要使用该序列的绝对值。然而,与具有正值和负值术语的序列合作是不同的。

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Always think of using the Root Test when a series has powers in the numerator and denominator such as with  $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{\cos (n \pi)}{n+2} \right) ^n\end{align*}$ . Let's use the $\begin{align*}n\end{align*}$ th Root Test, determine if the series $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{\cos (n \pi)}{n+2} \right) ^n\end{align*}$ is absolutely convergent, conditionally convergent, or divergent.
::当一个序列在分子和分母(如n=1(cosn+2)n)中拥有权力时,总是想着使用根测试。让我们使用nth根测试,确定序列 *n=1(cosn+2)n是否绝对集合,有条件集合,或差异。

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The $\begin{align*}n\end{align*}$ th Root Test for absolute convergence says to look at $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}\end{align*}$ .
::绝对趋同的Nth根测试 表示要看Limnnnan。

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With $\begin{align*}a_n=\left(\frac{\cos(n \pi)}{n+2} \right)^n\end{align*}$ , then
::用 an= (cosn+2n, 然后用 an=(cosn+2n)n, then

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$$\begin{align*}\lim_{n \to \infty} \sqrt[n]{|a_n|} &=\lim_{n \to \infty} \sqrt[n]{\bigg| \left(\frac{\cos(n \pi)}{n+2} \right)^n \bigg|} \\ &=\lim_{n \to \infty} \sqrt[n]{\bigg|\left(\frac{1}{n+2} \right)^n \bigg|} \\ &=\lim_{n \to \infty} \frac{1}{n+2} \\ &=0 < 1\end{align*}$$
::{\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! {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}不!

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By the $\begin{align*}n\end{align*}$ th Root Test for absolute convergence, the series  $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{\cos(n \pi)}{n+2} \right)^n\end{align*}$ is absolutely convergent.
::到 nth 根测试时, 绝对趋同, 序列 {n=1 (cosn+2)n 绝对趋同 。

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For some series, neither the Ratio Test nor the $\begin{align*}n\end{align*}$ th Root Test provide a conclusive indication of absolute convergence, and the $\begin{align*}n\end{align*}$ th term Divergence Test is also inconclusive. In this situation, other tests must be tried.
::对于某些系列,比率测试和 nth 根测试都没有提供绝对趋同的决定性迹象,Nth 术语“差异测试”也是没有结论的。在这种情况下,必须尝试其他测试。

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Let's use the Ratio or $\begin{align*}n\end{align*}$ th Root Test, determine if the series $\begin{align*}\sum\limits_{n=2}^{\infty}(-1)^{n-1} \left(\frac{n^2}{n^3-2} \right)\end{align*}$ is absolutely convergent, conditionally convergent, or divergent.
::让我们使用比率或 nth 根测试, 确定序列 *n=2 (- 1)n- 1 (n2n3-2) 是绝对趋同的, 有条件的趋同的, 还是不同的 。

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The Ratio Test for absolute convergence says to look at  $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n+1}{a_n} \bigg|\end{align*}$ .
::绝对趋同比率测试表示要查看 limnan+1an。

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With $\begin{align*}a_n=\left(\frac{n^2}{n^3-2} \right)\end{align*}$ , then
::然后用一个=(n2n3-2),然后

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$$\begin{align*}\lim_{n \to \infty} \bigg|\frac{a_{n+1}}{a_n} \bigg| &=\lim_{n \to \infty} \left[\frac{(n+1)^2}{(n+1)^3-2} \cdot \frac{n^3-2}{n^2} \right] \\ &=\lim_{n \to \infty} \left[\frac{n^5}{n^5} \frac{\left(1+ \frac{1}{n} \right)^2}{\left(1+ \frac{1}{n} \right)^3 - \frac{2}{n^3}} \cdot \frac{ \left(1- \frac{2}{n^3} \right)}{1} \right] \\ &=1\end{align*}$$
::[(n+1)(n+1)(n+1)3-2n3-2n2]=limn{[n5n5(1+1n)2(1+1n)3-2n3}(1-2n3)1=1

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The Ratio Test for absolute convergence is inconclusive.
::绝对趋同比率测试没有结论。

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If we try the $\begin{align*}n\end{align*}$ th Term Divergence Test: $\begin{align*}\lim\limits_{n \to \infty} \left(\frac{n^2}{n^3 -2} \right)=\lim\limits_{n \to \infty} \left(\frac{1}{n} \frac{1}{1- \frac{2}{n^3}} \right)=0\end{align*}$ . This result does not tell us anything about divergence!
::如果我们尝试 nn Term Divergence 测试 : limn( n2n3-2) =limn( 1n11- 2n3) =0。 此结果没有告诉我们任何差异 !

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The $\begin{align*}n\end{align*}$ th Root Test for absolute convergence also is inconclusive about absolute convergence. Try performing the test on your own to show $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{|a_n|}=1\end{align*}$ , by figuring out how to evaluate the limits:
::Nth 根测试对绝对趋同的绝对趋同也无定论。 尝试独立进行测试以显示 limnnan1, 方法为如何评估限制 :

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$$\begin{align*}\lim_{n \to \infty} \sqrt[n]{|a_n|} &=\lim_{n \to \infty} \sqrt[n]{\frac{n^2}{n^3 -2}} \\ &=\lim_{n \to \infty} \left(\frac{n^2}{n^3 -2} \right)^{\frac{1}{n}} \\ &=\left(\frac{\lim\limits_{n \to \infty} n^{\frac{2}{n}}}{\lim\limits_{n \to \infty}(n^3 -2)^{\frac{1}{n}}} \right) && \ldots \text{Can you evaluate these two limits}? \\ &=\frac{1}{1}=1\end{align*}$$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}你能评估这两个限制吗? {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}=11=1 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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Let’s try the Limit Comparison Test on the series $\begin{align*}\sum\limits_{n=2}^{\infty} \bigg|(-1)^{n-1} \left(\frac{n^2}{n^3 -2} \right) \bigg|\end{align*}$  and use  $\begin{align*}\sum\limits_{n=2}^{\infty} \frac{n^2}{n^3}=\sum\limits_{n=2}^{\infty} \frac{1}{n}\end{align*}$ as the comparison series.
::让我们试一下该序列的“限制比较测试” , 并使用“n=2n2n3-2” 来比较序列。

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Then $\begin{align*}\lim\limits_{n \to \infty} \frac{u_k}{v_k}=\lim\limits_{n \to \infty} \frac{\frac{n^2}{n^3 -2}}{\frac{1}{n}}\end{align*}$ , and
::然后是Limnukvk=limnn2n3 -21n,和

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$$\begin{align*}\lim_{n \to \infty} \frac{\frac{n^2}{n^3 -2}}{\frac{1}{n}} &=\lim_{n \to \infty} \frac{n^3}{n^3} \frac{1}{1- \frac{2}{n^3}} \\ &=1 > 0\end{align*}$$
::立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺 立方公尺

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Since $\begin{align*}\sum\limits_{n=2}^{\infty} \frac{1}{n}\end{align*}$  is a harmonic series and diverges , $\begin{align*}\sum\limits_{n=2}^{\infty} \bigg|(-1)^{n-1} \left(\frac{n^2}{n^3 -2} \right) \bigg|\end{align*}$  also diverges absolutely.
::@n=21n是一个调和序列和差异, @n=2_(-1)_(1)_(n2n3-2)_(2)_也绝对不同。

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To test for conditional convergence, we can use the .
::为了测试有条件的趋同,我们可以使用它。

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The term $\begin{align*}\left(\frac{n^2}{n^3 -2} \right)\end{align*}$  is always decreasing, and $\begin{align*}\lim\limits_{N \to \infty} \left(\frac{n^2}{n^3 -2} \right)=0\end{align*}$ .
::该术语(n2n3-2) 总是在减少,而limN(n2n3-2)=0。

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Therefore, the series $\begin{align*}\sum\limits_{n=2}^{\infty} (-1)^{n-1} \left(\frac{n^2}{n^3 -2} \right)\end{align*}$  is conditionally convergent!
::因此,该系列=2(-1)n-1(n2n3-2)是有条件的趋同!

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked what the Ratio and Root Tests for $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg|3 \left(- \frac{2}{3} \right)^{n-1} \bigg|\end{align*}$  tell you about its absolute convergence. 
::更早之前,有人问您,“比率和根测试”1=13 (-23)-1 告诉您,它绝对一致。

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To test the series  $\begin{align*}\sum\limits_{n=1}^{\infty} 3 \left(- \frac{2}{3} \right)^{n-1}\end{align*}$ for absolute convergence we create the series $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg|3 \left(- \frac{2}{3} \right)^{n-1} \bigg|= \sum\limits_{n=1}^{\infty} 3 \left(\frac{2}{3} \right)^{n-1}\end{align*}$  which we know is a convergent geometric series with $\begin{align*}r=\frac{2}{3} < 1\end{align*}$ . Do the Ratio and Root Tests show this?
::为测试 {n= 13(- 23n- 1) 序列绝对趋同性, 我们创建此序列 {n= 1}3}(- 23n- 1) = 13( 23)n- 1) 我们知道这是一个共和几何序列, r= 23<1. 比率和根测试显示这一点吗 ?

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For the Ratio Test: $\begin{align*}\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n}=\lim\limits_{n \to \infty} \frac{3 \left(\frac{2}{3} \right)^n}{3 \left(\frac{2}{3} \right)^{n-1}}=\lim\limits_{n \to \infty} \frac{3 \left(\frac{2}{3} \right)^n}{3 \left(\frac{2}{3} \right)^{n-1}}=\frac{2}{3} < 1\end{align*}$ . The series  $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg|3 \left(- \frac{2}{3} \right)^{n-1} \bigg|=\sum\limits_{n=1}^{\infty} 3 \left(\frac{2}{3} \right)^{n-1}\end{align*}$ converges absolutely by the Ratio Test, and therefore  $\begin{align*}\sum\limits_{n=1}^{\infty} 3 \left(- \frac{2}{3} \right)^{n-1}\end{align*}$ converges.
::比率测试: limnan+1=limn3(23)n3(23)n3(23)_1=23 <1. 该序列=13(23)n-1=13(23)n=13(23)n-1通过比率测试绝对趋同, 因此n=13(23)n- 1趋同 。

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For the Root Test: 

$$\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{a_n} &=\lim\limits_{n \to \infty} \sqrt[n]{3 \left(\frac{2}{3} \right)^{n-1}} \\ &=\lim\limits_{n \to \infty} 3^{\frac{1}{n}} \left(\frac{2}{3} \right)^{\frac{(n-1)}{n}}=\lim\limits_{n \to \infty} 3^{\frac{1}{n}} \cdot \lim\limits_{n \to \infty} \left(\frac{2}{3} \right)^{\frac{(n-1)}{n}}=1 \cdot \left(\frac{2}{3} \right) < 1\end{align*}$$
::根测试: limnnan=limnn}33(23)n-1=limn31n(23)n=limn31n(23)n=limn31n(23)n=1}<1

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The series  $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg|3 \left(- \frac{2}{3} \right)^{n-1} \bigg|=\sum\limits_{n=1}^{\infty} 3 \left(\frac{2}{3} \right)^{n-1}\end{align*}$ converges absolutely by the Root Test, and therefore  $\begin{align*}\sum\limits_{n=1}^{\infty} 3 \left(- \frac{2}{3} \right)^{n-1}\end{align*}$ converges.
::序列=13 (- 23n- 11=13( 23n-1) 在根测试时绝对一致, 因此n=13(- 23n-1) 趋同 。

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The application of the Ratio and Root Tests to the series to show absolute convergence was based on taking the absolute value of all terms. Therefore the modifications to the two tests to make each applicable to all series are:
::将“比率和根测试”应用于该系列,以显示绝对趋同,其依据是所有术语的绝对值。

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$\begin{align*}\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_n} \Rightarrow \lim\limits_{n \to \infty} \bigg|\frac{a_{n+1}}{a_n} \bigg|\end{align*}$  and $\begin{align*}\lim\limits_{n \to \infty} \sqrt[n]{a_n} \Rightarrow \lim\limits_{n \to \infty} \sqrt[n]{|a_n|}\end{align*}$ .
::

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Example 2
::例2

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Using the Ratio Test, determine if the series $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{7n+3}\end{align*}$  is absolutely convergent, conditionally convergent, or divergent. If the Ratio Test is inconclusive, try other tests.
::使用“比率测试”来确定“%n=1(- 1)n7n+3”序列是否绝对趋同、有条件趋同或不同。如果“比率测试”没有结果,请尝试其他测试。

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The Ratio Test for absolute convergence says to look at $\begin{align*}\lim\limits_{n \to \infty} \bigg|\frac{a_n +1}{a_n} \bigg|\end{align*}$ .
::绝对趋同比率测试表示要查看 limnan+1an。

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With $\begin{align*}a_n=\frac{(-1)^n}{7n+3}\end{align*}$ , then
::使用 an= (- 1) n7n+3, 然后

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lim n → ∞ | a n + 1 a n | = lim n → ∞ | ( − 1 ) n + 1 7 ( n + 1 ) + 3 ( − 1 ) n 7 n + 3 | = lim n → ∞ | 7 n + 3 7 n + 10 | = lim n → ∞ | 7 n ( 1 + 3 7 n ) 7 n ( 1 + 10 7 n ) | = 1

$$\begin{align*}\lim_{n \to \infty} \bigg|\frac{a_{n+1}}{a_n} \bigg| &=\lim_{n \to \infty} \bigg|\frac{\frac{(-1)^{n+1}}{7(n+1)+3}}{\frac{(-1)^n}{7n+3}} \bigg| \\ &=\lim_{n \to \infty} \bigg|\frac{7n+3}{7n+10} \bigg| \\ &=\lim_{n \to \infty} \bigg|\frac{7n \left(1+ \frac{3}{7n} \right)}{7n \left(1+ \frac{10}{7n} \right)} \bigg| \\ &=1\end{align*}$$
::=================================================================================================================================================== ====================================================================================================== ==================================================================================================================================================================================================================================================================

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The Ratio Test is inconclusive on the absolute convergence of the series. Another test must be tried.
::比率测试不能确定序列绝对趋同。 必须尝试另一个测试 。

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Let’s see if the absolute series diverges. If the $\begin{align*}n\end{align*}$ th Term Divergence Test is used the result is: $\begin{align*}=\lim\limits_{n \to \infty} \bigg|\frac{1}{7n+3} \bigg|=0\end{align*}$ , which is inconclusive in establishing divergence. Another test must be tried.
::让我们看看绝对序列是否有差异。 如果使用 nn Terral divergence 测试的结果是 : =limn17n+30, 在确定差异方面没有结果。 必须尝试另一个测试 。

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If the $\begin{align*}n\end{align*}$ th Root Test is used the result is:
::如果使用 nth 根测试,结果为:

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$$\begin{align*}\lim_{n \to \infty} \sqrt[n]{|a_n|} &=\lim_{n \to \infty} \sqrt[n]{\bigg|\frac{(-1)^n}{7n+3} \bigg|} \\ &=\lim_{n \to \infty} \frac{1}{(7n+3)^{\frac{1}{n}}} \\ &=\frac{1}{\lim\limits_{n \to \infty} (7n+3)^{\frac{1}{n}}} && \ldots \text{To evaluate} \lim_{n \to \infty}(7n+3)^{\frac{1}{n}},\ \text{let} \ y=(7n+3)^{\frac{1}{n}}. \\ &&& \text{Then} \ \lim_{n \to \infty} y \ \text{is related to} \ \lim_{n \to \infty}(\ln y) \ \text{as follows}: \\ \lim_{n \to \infty} (\ln y) &=\lim_{n \to \infty} [\ln(7n+3)^{\frac{1}{n}}] \\ &=\lim_{n \to \infty} \left[\frac{\ln (7n+3)}{n} \right] && \ldots \text{Looks like} \ \frac{\infty}{\infty}. \\ &=\lim_{n \to \infty} \left[\frac{\frac{7}{(7n+3)}}{1} \right] && \ldots \text{Using L’Hopital’s Rule}. \\ &=0. &&\ldots \lim_{n \to \infty} (\ln y)=0 \ \text{means} \ \lim_{n \to \infty} y=e^0=1=\lim_{n \to \infty} (7n+3)^{\frac{1}{n}}. \\ \lim_{n \to \infty} \sqrt[n]{\bigg|\frac{(-1)^n}{7n+3} \bigg|} &=1\end{align*}$$
::===================================================================================================================================================================================

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The Root Test is also not conclusive on absolute convergence.
::" 根测试 " 也不是绝对趋同的定论。

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Let’s try the on $\begin{align*}\sum\limits_{n=1}^{\infty} \bigg|\frac{(-1)^n}{7n+3} \bigg|\end{align*}$  to test for absolute convergence.
::让我们试一下在 @n=1(- 1)n7n+3上测试绝对趋同 。

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Let $\begin{align*}f(x)=\frac{1}{7x+3}\end{align*}$ , $\begin{align*}x \ge 1\end{align*}$ . Then, $\begin{align*}f(x)\end{align*}$  is positive, continuous, and decreasing thus satisfying the conditions for using the Integral Test.
::Let f(x) = 17x+3, x%1. f(x) 是正数、连续数和递减数,从而满足了使用综合测试的条件。

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$$\begin{align*}\int_{1}^{\infty} \frac{1}{7x+3}dx &=\lim_{p \to \infty} \int_{1}^{p} \frac{1}{7x+3}dx \\ &=\lim_{p \to \infty} \left[\frac{1}{7} \ln (7p+3)- \frac{1}{7} \ln (7+3) \right] \\ &=\infty\end{align*}$$
::117x+3dx = limpp117x+3dx = limp[17ln(7p+3) -17n(7+3)]\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\7\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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By the Integral Test, the series does not converge absolutely, it diverges.
::通过综合测试,系列不是绝对趋同的,而是不同的。

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Now, since $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{7n+3}\end{align*}$  is an , use the Alternating Series Test to see if there is conditional convergence.
::现在,既然=1(-1)n7n+3是一个,请使用交替序列测试,看看是否有有条件的趋同。

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Since  $\begin{align*}\frac{1}{7n+3}\end{align*}$ is decreasing and $\begin{align*}\lim\limits_{n \to \infty} \frac{1}{7n+3}=0\end{align*}$ , the series converges.
::由于17n+3正在下降 和limn17n+3=0, 系列会合。

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The series $\begin{align*}\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{7n+3}\end{align*}$  is conditionally convergent!
::#n=1(- 1) n7n+3 序列是有条件合并的 !

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Review
::回顾

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For #1-7, determine whether the  series is absolutely convergent, conditionally convergent, or divergent by using the Ratio Test. If the Ratio Test is inconclusive, try another test:
::对于 # 1-7, 确定该序列是否绝对趋同, 有条件趋同, 或者通过使用比率测试而出现差异。 如果比率测试没有结果, 请尝试另一个测试 :

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1. $\begin{align*}\sum\limits_{k=1}^{\infty} (-1)^k \frac{1}{2^{k-1}}\end{align*}$
   ::*k=1 (- 1) k12k- 1
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2. $\begin{align*}\sum\limits_{k=1}^{\infty} \frac{(-1)^k}{k !}\end{align*}$
   ::K=1(-1)kk!
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3. $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{(2n)!}{(-4)^n}\end{align*}$
   ::N=1( 2n)! (- 4) n
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4. $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{(-1)^n (1.5)^n}{n^4}\end{align*}$
   ::n=1(-1)n(1.5)nn4
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5. $\begin{align*}\sum\limits_{k=1}^{\infty} \frac{(-1)^{k+1}}{2k+1}\end{align*}$
   ::*\ k=1 (- 1) k+12k+1
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6. $\begin{align*}\sum\limits_{n=1}^{\infty} (-1)^n \frac{n^2}{2^n}\end{align*}$
   ::*n=1(-1)nn22n
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7. $\begin{align*}\sum\limits_{n=1}^{\infty}(-1)^n \frac{3^{n-1}}{n^4}\end{align*}$
   ::*n=1(- 1)n3n-1n4

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For #8-15, determine whether the  series is absolutely convergent, conditionally convergent, or divergent by using the $\begin{align*}n\end{align*}$ th Root Test. If the test is inconclusive, try another test:
::对于# 8-15, 确定序列是否绝对趋同, 有条件趋同, 或者通过使用 nth 根测试而出现差异。 如果测试没有结果, 请尝试另一个测试 :

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8. $\begin{align*}\sum\limits_{n=1}^{\infty} (-1)^n \frac{n+1}{3^n} \end{align*}$
   ::*n=1(-1)n+13n
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9. $\begin{align*}\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{[\ln(n+1)]^n}\end{align*}$
   ::*n=1(- 1)n [n(n+1)]n
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10. $\begin{align*}\sum\limits_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^2 -1}\end{align*}$
    ::*n=1(-1)n+1nn2-1
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11. $\begin{align*}\sum\limits_{n=2}^{\infty} \frac{(-1)^n}{\ln(2n)}\end{align*}$
    ::*n=2(- 1)nln( 2n)
12. $\begin{align*}\frac{2}{1.3}- \left(\frac{3}{2.4} \right)^2+ \left(\frac{4}{3.5} \right)^3- \left(\frac{5}{4.6} \right)^4+ \cdots\end{align*}$
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13. $\begin{align*}\sum\limits_{n=1}^{\infty} (-1)^n \left(\frac{4n+2}{19n-1} \right)^{2n}\end{align*}$
    ::*n=1(- 1)n( 4n+219n- 1)n2n
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14. $\begin{align*}\sum\limits_{n=1}^{\infty} (-1)^n \left(\frac{n^2 -1}{n^2 +1} \right)^{n}\end{align*}$
    ::*n=1(- 1)n(n2- 1n2+1)n
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15. $\begin{align*}\sum\limits_{n=1}^{\infty} \left(\frac{(-1)^n \ln n}{n} \right)^n\end{align*}$
    ::n=1(- 1) nlnnnnn

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Review (Answers)
::回顾(答复)

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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 。