高數(shù)重要極限證明原創(chuàng)中英文對(duì)照版

高數(shù)重要極限證明原創(chuàng)中英文對(duì)照版

重要極限

Important Limit

作者 趙天宇

Author:Panda Zhao

我今天想在這里證明高等數(shù)學(xué)中的一個(gè)重要極限：

Today I want to prove animportant limit of higher mathematics by myself:

想要證明上述極限，我們先要去證明一個(gè)數(shù)列極限：

If we want to give evidence ofthe limit, first of all, there are a limit of a series of numbers according toa certain rule we need to certify:

想要證明這個(gè)極限，我首先要介紹一個(gè)定理和一個(gè)法則：

Before we begin to prove thelimit, there are one theorem and one rule that are the key point we need to introduce:

1.       牛頓二項(xiàng)式定理（Binomialtheorem）

定理的定義為：

Definition of Binomial theorem:

其中 ，稱為二項(xiàng)式系數(shù)，又有 的記法。

Among the formula: we define the  as binomialcoefficient, it can be remembered to.

牛頓二項(xiàng)式定理（Binomial theorem）驗(yàn)證和推理過(guò)程：

The process of the ratiocination of Binomialtheorem:

采用數(shù)學(xué)歸納法

We consider to use the mathematical inductionto solve this problem.

當(dāng)n = 1時(shí)(While n = 1:),

;

假設(shè)二項(xiàng)展開式在n=m時(shí)成立。

We can make a hypothesis that the binomial expansionequation is true when n = m.

設(shè)n=m+1，則：So if we suppose that n equal mplus one, we will CONTINUE to deduce:
 

具體步驟解釋如下：

The specific step of interpretation :

第三行：將a、b乘入；

The 3rd line: a and b are multiplied into the binomial expansion equation.;

第四行：取出k=0的項(xiàng)；

The 4th line: take out of theitem which includes the k = 0 in the binomial expansion equation.;

第五行：設(shè)j=k-1；

The 5th line: making a hypothesisthat is j = k-1;

第六行：取出k=m+1項(xiàng)；

The 6th line: What we need totake out of the item including k=m+1 in the binomial expansion equation.

第七行：兩項(xiàng)合并；

The 7th line: Combining the twobinomial expansion equation.

第八行：套用帕斯卡法則；

The 8th line: At this line weneed to use the Pascal’s Rule to combine the binomial expansion equation whichare
.; 

接下來(lái)介紹一下帕斯卡法則(Pascal’s Rule)。

So at this moment, we should get someknowledge about what the Pascal’s Rule is. Let’s see something about it:

帕斯卡法則(Pascal’s Rule)：組合數(shù)學(xué)中的二項(xiàng)式系數(shù)恒等式，對(duì)于正整數(shù)n、k(k<=n)有：

Pascal’s Rule: a binomial coefficientidentical equation of combinatorial mathematics. For the positive integer n andk (k<=n), there is a conclusion:

                  通常也可以寫成：
                  There is also commonly written:

代數(shù)證明：

Algebraic proof:

重寫左邊：

We can rewrite the left combinatorial item:

通分；reductionof fractions to a common.

                                         合并多項(xiàng)式；combining the polynomial.

                          證明完成；The Pascal’s Rule has been proved.

接下來(lái)只要要證明是單調(diào)增加并且有界的，那么就可以得到它存在極限，我們通常稱它的極限為e。

So what is our next step? The progression ofnumbers according to a certain rule of should be proved that it is a monotonicincrease sequence and has a limitation. If we can do these things, we will drawa conclusion that the sequence has an limitation which we generally call e.

類似的，我們可以得到：

We can analogously get the:

可見，  和相比，除了前兩個(gè)1相等之外，后面的項(xiàng)都要小，并且多一個(gè)值大于0的項(xiàng)目，因此：

Thus it can be seen, comparing   with  , all of the items of the   are lower thanthese items in  except the 1stand the 2rd one are equaling. In addition it has an item whose value is biggerthan zero that is in the . So we can get a point :

所以數(shù)列是單調(diào)遞增的得證，接下來(lái)證明其有界性：

Because of the point, we can prove thesequence is an monotonic increase sequence, so we remain only one thing shouldbe proved that is the sequence’s limitation. So let’s get it :

可見{  }是有界的，所以根據(jù)數(shù)列極限存在準(zhǔn)則可得：

Thus it can be seen , the sequence of  has a limitation , as we know, we can draw aconclusion by the means of the rule of limitation of sequence exiting: