Course: 11.11 部分总和 | The Way To Learn

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部分总和
In the 18th century, mathematician Leonhard Euler solved one of the foremost infinite series problems of his day by examining the series $\begin{align*}\sum \limits_{n=1}^{\infty}\frac{2}{n(n+1)}\end{align*}$ .
::在18世纪,数学家Leonhard Euler通过审查系列丛书"n=12n(n+1),解决了他这一天最无穷无尽的系列问题之一。

Find the first five of this series and make an observation about the sum of the infinite series.
::找到这个系列的前五个并观察无限序列的总和

Source: Plus Magazine
::资料来源:《加杂志》。

Partial Sums
::部分总和

An infinite series is a series with an infinite number of terms. In other words, the value of $\begin{align*}n\end{align*}$ increases without bound as shown in the series below.
::无限系列是一系列,用无限数的术语。换句话说,n的增加值没有约束,如下文系列所示。

$\begin{aligned} \sum_{n = 1}^{\infty} 3 n + 1 & = 4 + 7 + 10 + 13 + \dots \\ \sum_{n = 1}^{\infty} 4 (2)^{n - 1} & = 4 + 8 + 16 + 32 + \dots \\ \sum_{n = 1}^{\infty} 8 {(\frac{1}{2})}^{n - 1} & = 8 + 4 + 2 + 1 + \frac{1}{2} + \dots \end{aligned}$
$\begin{align*}\sum \limits_{n=1}^{\infty}3n+1 & =4+7+10+13+ \ldots \\ \sum \limits_{n=1}^{\infty}4(2)^{n-1} & =4+8+16+32+ \ldots\\ \sum \limits_{n=1}^{\infty}8 \left(\frac{1}{2}\right)^{n-1}&=8+4+2+1+ \frac{1}{2}+ \ldots\end{align*}$
::=13n+1=4+7+10+13+...n=14(2)n-1=4+8+16+32+...n=18(12)n-1=8+4+2+1+12+...

These sums continue forever and can increase without bound.
::这些数额将永远持续下去,并且可以无拘无束地增加。

Since we cannot find the sums of these series by adding all the terms, we can analyze their behavior by observing patterns within their partial sums . A partial sum is a sum of a finite number of terms in the series. We can look at a series of these sums to observe the behavior of the infinite sum. Each of these partial sums is denoted by $\begin{align*}S_n\end{align*}$ where $\begin{align*}n\end{align*}$ denotes the index of the last term in the sum. For example, $\begin{align*}S_6\end{align*}$ is the sum of the first 6 terms in an infinite series.
::由于我们无法通过添加所有术语来找到这些序列的总和, 我们可以通过观察其部分总和中的规律来分析它们的行为。部分总和是一系列中数量有限的术语的总和。我们可以看一系列这些总和来观察无限总和的行为。这些部分总和中的每一部分都由 Sn 表示, 其中n 表示最后一个术语的指数。例如, S6 是无限序列中前六个术语的总和。

Let's find the first five partial sums of $\begin{align*}\sum \limits_{n=1}^{\infty} 2n-1\end{align*}$ and make an observation about the sum of the infinite series.
::让我们先找到前五个部分总和的%n=12n-1, 并对无限序列的总和进行观察。

The first five partial sums are $\begin{align*}S_1, S_2, S_3, S_4\end{align*}$ and $\begin{align*}S_5\end{align*}$ . To find each of these sums we will need the first five terms of the : 1, 3, 5, 7, 9. No we can find the partial sums as shown:
::头五个部分数额是S1、S2、S3、S4和S5。为了找到其中每一个数额,我们需要前五个条件:1、3、5、7、9。

$\begin{aligned} S_{1} & = a_{1} = 1 \\ S_{2} & = a_{1} + a_{2} = 1 + 3 = 4 \\ S_{3} & = a_{1} + a_{2} + a_{3} = 1 + 3 + 5 = 9 \\ S_{4} & = a_{1} + a_{2} + a_{3} + a_{4} = 1 + 3 + 5 + 7 = 16 \\ S_{5} & = a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 1 + 3 + 5 + 7 + 9 = 25 \end{aligned}$
$\begin{align*}S_1&=a_1=1 \\ S_2&=a_1+a_2=1+3=4 \\ S_3&=a_1+a_2+a_3=1+3+5=9 \\ S_4&=a_1+a_2+a_3+a_4=1+3+5+7=16 \\ S_5&=a_1+a_2+a_3+a_4+a_5=1+3+5+7+9=25\end{align*}$
::S1=a1=1S2=1S2=a1+a2=1+a2+3=1+3=3=4S3=a1+a2+3+3+5=9S4=a1+a2+a3+a4=1+3+3+5+7=16S5=a1+a2+a3+a3+a4+a5=1+3+5+5+9=25

Notice that each sum can also be found by adding the $\begin{align*}n^{th}\end{align*}$ term to the previous sum: $\begin{align*}S_n=S_{n-1}+a_n\end{align*}$ .
::通知中说,每笔金额还可以通过在前一笔金额(Sn=Sn-1+an)中加上n期来找到。

For example: $\begin{align*}S_5=S_4+a_4=16+9=25\end{align*}$
::例如:S5=S4+a4=16+9=25

The sequence of the first five partial sums is 1, 4, 9, 16, 25. This pattern will continue and the terms will continue to grow without bound. In other words, the partial sums continue to grow and the infinite sum cannot be determined as it is infinitely large.
::前五个部分金额的顺序是1、4、9、16、25。这种模式将继续下去,术语将无约束地继续增长。换句话说,部分金额继续增长,无法确定无限金额,因为它是无限大的。

Now, let's find the first five partial sums of $\begin{align*}\sum \limits_{n=1}^{\infty} \left(\frac{1}{2}\right)^{n-1}\end{align*}$ and make an observation about the sum of the infinite series.
::现在,让我们先找到前五个部分总和的 'n=1\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\可以观察无限序列的总和。

The first five terms of this sequence are: $\begin{align*}1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{16}\end{align*}$ . The partial sums are thus:
::这一顺序的前五个条件为:1,12,14,18,116。

$\begin{aligned} S_{1} & = 1 \\ S_{2} & = 1.5 \\ S_{3} & = 1.75 \\ S_{4} & = 1.875 \\ S_{5} & = 1.9375 \end{aligned}$
$\begin{align*}S_1&=1 \\ S_2&=1.5 \\ S_3&=1.75 \\ S_4&=1.875 \\ S_5&=1.9375\end{align*}$
::S1=1S2=1.5S3=1.75S4=1.875S5=1.9375

Consider what happens with each subsequent term: We start with 1 and add $\begin{align*}\frac{1}{2}\end{align*}$ putting us halfway between 1 and 2. Then we add $\begin{align*}\frac{1}{4}\end{align*}$ , putting us halfway between 1.5 and 2. Each time we add another term, we are cutting the distance between our current sum and 2 in half. If this pattern is continued, we will get ever closer to 2 but never actually reach two. Therefore, the sum is said to “ converge to” or “approach” 2.
::想想接下来的每个任期会怎样:我们从1开始,加上12个,让我们在1到2之间走一半,然后我们加上14个,让我们在1.5到2之间走一半,每当我们增加另一个任期,我们就会缩短我们目前和2之间的距离。如果这种模式继续下去,我们就会更接近2个,但实际上永远不会达到2个。因此,这个数额被说成是“通向”或“接近”2个。

To support our conjecture further, we can use the calculator to find the $\begin{align*}50^{th}\end{align*}$ partial sum: $\begin{align*}S_{50}=2\end{align*}$ . Eventually, if we sum enough terms, the calculator will give us the value to which the sum approaches due to rounding.
::为了进一步支持我们的推测,我们可以使用计算器来找到第50部分总和:S50=2。最终,如果我们计算出足够的条件,计算器将给我们四舍五入的总和方法的价值。

Finally, let's find the first five partial sums of $\begin{align*}\sum \limits_{n=1}^{\infty} \frac{1}{n}\end{align*}$ , the “harmonic series” and make an observation about the sum of the infinite series. (You may need to find addition partial sums to see the behavior of the infinite series.)
::最后,让我们先找到前五个部分总和 n=1%1n, “ 调音序列” , 并对无限序列的总和进行观察。 (你可能需要找到额外的部分总和才能看到无限序列的行为。 )

Use the calculator to find the following sums:
::使用计算器查找以下金额:

$\begin{aligned} S_{1} & = s u m (s e q (1 / x, x, 1, 1)) = 1 \\ S_{2} & = s u m (s e q (1 / x, x, 1, 2)) = 1.5 \\ S_{3} & = s u m (s e q (1 / x, x, 1, 3)) = 1.833 \\ S_{4} & = s u m (s e q (1 / x, x, 1, 4)) = 2.083 \\ S_{5} & = s u m (s e q (1 / x, x, 1, 5)) = 2.283 \end{aligned}$
$\begin{align*}S_1&=sum(seq(1/x,x,1,1))=1 \\ S_2&=sum(seq(1/x,x,1,2))=1.5 \\ S_3&=sum(seq(1/x,x,1,3))=1.833 \\ S_4&=sum(seq(1/x,x,1,4))=2.083 \\ S_5&=sum(seq(1/x,x,1,5))=2.283\end{align*}$
::S1=Ssum(1,x,x,1,1)=1,S2=sum(seq(1/x,x,1,2))=1.5S3=sum(seq(1/x,x,x,1,3))=1.833S4=sum(seq(1/x,x,x,1,4))=2.083S5=sum(seq(1,x,x,1,5)=2.283

In this series, the behavior is not quite as clear. Consider some additional partial sums:
::在这一系列行为中,行为并不十分明确。

$\begin{aligned} S_{50} & = 4.499 \\ S_{100} & = 5.187 \\ S_{500} & = 6.793 \end{aligned}$
$\begin{align*}S_{50}&=4.499 \\ S_{100}&=5.187 \\ S_{500}&=6.793\end{align*}$
::S50=4.4999S100=5.187S500=6.793

In this case, the partial sums don’t seem to have a bound. They will continue to grow and therefore there is no finite sum.
::在这种情况下,部分金额似乎没有约束性。它们将继续增长,因此没有限定金额。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the first five partial sums of the series $\begin{align*}\sum \limits_{n=1}^{\infty}\frac{2}{n(n+1)}\end{align*}$ , and make an observation about the sum of the infinite series.
::更早之前,有人要求你找到 =12n(n+1)系列的前五个部分总和, 并对无限序列的总和进行观察。

The first five terms of this sequence are: $\begin{align*}1, \frac{1}{3}, \frac{1}{6}, \frac{1}{10}, \frac{1}{15}\end{align*}$ . The partial sums are thus:
::这一顺序的前五个条件为:1,13,16,110,115。

$\begin{aligned} S_{1} & = 1 \\ S_{2} & = 1.333 \\ S_{3} & = 1.5 \\ S_{4} & = 1.6 \\ S_{5} & = 1.6666 \end{aligned}$
$\begin{align*}S_1&=1 \\ S_2&=1.333 \\ S_3&=1.5 \\ S_4&=1.6 \\ S_5&=1.6666\end{align*}$
::S1=1S2=1.333S3=1.5S4=1.6S5=1.666

If this pattern is continued, we will get ever closer to 2 but never actually reach two. Therefore, the sum is said to “converge to” or “approach” 2.
::如果这种模式继续下去,我们将更接近于2,但实际上永远不会达到2,因此,这笔金额据说是“汇合”或“近似”2。

Find the first five partial sums of the infinite series below and additional partial sums if needed to determine the behavior of the infinite series. Use the calculator to find the partial sums.
::查找下方无限序列的前五个部分总和, 如果确定无限序列的行为需要额外部分总和。使用计算器查找部分总和。

Example 2
::例2

$\begin{align*}\sum \limits_{n=1}^{\infty} 4 \left(\frac{3}{2}\right)^{n-1}\end{align*}$
::=14(321)

$\begin{align*}S_1=4; \ S_2=10; \ S_3=19; \ S_4=32.5; \ S_5=52.75;\end{align*}$ The partial sums are growing with increasing speed and thus the infinite series will have no bound.
::S1=4;S2=10;S3=19;S4=32.5;S5=52.75;部分金额正在以不断加快的速度增长,因此无限序列没有约束。

Example 3
::例3

$\begin{align*}\sum \limits_{n=1}^{\infty} 500 \left(\frac{2}{3}\right)^{n-1}\end{align*}$
::=1500(23)-1

$\begin{align*}S_1=500; \ S_2=833.333; \ S_3=1055.556; \ S_4=1203.704; \ S_5=1302.469;\end{align*}$ Here the sums seems to be growing by smaller amounts each time. Look at the sum additional partial sums to see if there is an apparent upper limit to their growth. $\begin{align*}S_{50}=1499.9999 \ldots=1500; \ S_{100}=1500\end{align*}$ . The sum is clearly approaching 1500 and thus the infinite series has a finite sum.
::S1=500;S2=833.333;S3=1055.556;S4=1203.704;S5=1302.469;这里的金额似乎每次增长较小。看看额外的部分金额,看其增长是否有明显的上限。S50=14999.9999999...=1500;S100=1500。这个金额显然接近1500,因此无限序列有一定的金额。

Example 4
::例4

$\begin{align*}\sum \limits_{n=1}^{\infty} \frac{5}{6n}\end{align*}$
::=1 56n

$\begin{align*}S_1=0.833; \ S_2=1.25; \ S_3=1.528; \ S_4=1.736; \ S_5=1.903;\end{align*}$ This sequence of partial sums is growing slowly, but will it approach a finite value or continue to grow? Look at additional partial sums: $\begin{align*}S_{50}=3.749; \ S_{100}=4.323; \ S_{500}=5.661\end{align*}$ . In this case, the sums continue to grow without bound so the infinite series will have no bound.
::S1=0.833;S2=1.25;S3=1.528;S4=1.736;S5=1.903;这一部分数额的顺序正在缓慢增长,但它会接近一定值还是继续增长?看看额外的部分数额:S50=3.749;S100=4.323;S500=5.661。在这种情况下,这些数额在不受约束的情况下继续增长,因此无限序列将没有约束。

Review
::回顾

Find the first five partial sums and additional partial sums as needed to discuss the behavior of each infinite series. Use your calculator to find the partial sums.
::查找前五个部分总和和和额外的部分总和, 以讨论每个无限序列的行为。使用您的计算器来找到部分总和。

$\begin{align*}\sum \limits_{n=1}^{\infty}5 \left(\frac{1}{2} \right)^{n-1}\end{align*}$
::=1 =5( 12) -1

$\begin{align*}\sum \limits_{n=1}^{\infty}2 \left(\frac{3}{4} \right)^{n-1}\end{align*}$
::=1 =2 =2 =3-1

$\begin{align*}\sum \limits_{n=1}^{\infty}10(0.9)^{n-1}\end{align*}$
::=110(0.9.9)n-1

$\begin{align*}\sum \limits_{n=1}^{\infty}8(1.03)^{n-1}\end{align*}$
::=1 =8 (1. 03n- 1)

$\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{2}n\end{align*}$
::=1 =1 =12n

$\begin{align*}\sum \limits_{n=1}^{\infty}\frac{10}{n}\end{align*}$
::=1 10n

$\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{2} \left(\frac{3}{4}\right)^{n-1}\end{align*}$
::=1=1=12(34n-1)

$\begin{align*}\sum \limits_{n=1}^{\infty}\frac{1}{n^2}\end{align*}$
::=1 1n2

$\begin{align*}\sum \limits_{n=1}^{\infty}6(0.1)^{n-1}\end{align*}$
::=1 =6(0.1)n-1

$\begin{align*}\sum \limits_{n=1}^{\infty}0.01n+5\end{align*}$
::=1 = 0.01n+5

$\begin{align*}\sum \limits_{n=1}^{\infty}2 \left(\frac{7}{8}\right)^{n-1}\end{align*}$
::=1=1 (2(78)-1

Which of the series above are arithmetic? Do any of them have a finite sum? Can you explain why?
::上面哪个序列是算术?它们中是否有一个有定数的?你能解释一下为什么吗?

Which of the series above are geometric? Do any of them have a finite sum? Can you explain why?
::上面哪个系列是几何级数?

What is the difference between the series that converge and diverge?
::趋同和相异的系列之间有什么区别?

Make a conjecture about the converging series in this problem set.
::推测一下这组问题中相融合的序列。

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

General

部分总和

Partial Sums ::部分总和

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Partial Sums
::部分总和

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)