Calculation of Zeno paradox

Nothing complicated , You should know all about advanced mathematics , Just a little bit , It's a kind of rehabilitation .

background

Zeno's paradox is simply expressed as , hypothesis A and B Distance between 1, Go alone $\frac{1}{2}$, Then I walked the rest of the way $\frac{1}{2}$, It's so infinite , Can this man go over there .

Middle school calculations

The first step is half done , The second step is to take half of the half , And so on , Add up every step of the journey to the total journey of this person , namely ：
$$L=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$$
Obviously, this is an equal ratio sequence , First item $a_1=\frac{1}{2}$, Gongbi $r=\frac{1}{2}$, General formula $a_n=(\frac{1}{2}{)}^n$. front n Items and are ：
$$S_n=a_1\frac{1-r^n}{1-r}=\frac{1}{2}\cdot \frac{1-(\frac{1}{2}{)}^n}{1-\frac{1}{2}}$$
So when n Towards infinity , Just figure out the front n The sum of the terms , We can judge whether this person can finish walking .

Calculation of higher numbers

n Before reaching infinity n The value of the sum of the terms is written like this ：
$$\underset{n\to +\infty }{\lim }{S}_n=\underset{n\to +\infty }{\lim }\frac{1}{2}\cdot \frac{1-(\frac{1}{2}{)}^n}{1-\frac{1}{2}}$$
The number of items in a sequence n It must be greater than zero , So it tends to be positive infinity . Because common ratio $r<1$, therefore $(\frac{1}{2}{)}^n$ Tends to infinity . Although the results can be seen at a glance , But for water , Here, I'd like to dismantle the upper formula first ：
$$\underset{n\to +\infty }{\lim }\frac{1}{2}\cdot (\frac{1}{1-\frac{1}{2}}-\frac{(\frac{1}{2}{)}^n}{1-\frac{1}{2}})$$
because $(\frac{1}{2}{)}^n$ Tends to infinity , That is, tend to 0, And the denominator is $1-\frac{1}{2}=\frac{1}{2}$, Such a comparison , The result is 0, Because the numerator is too small compared to the denominator , This ratio cannot be any ratio 0 Large number . The original formula becomes ：
$$\underset{n\to +\infty }{\lim }\frac{1}{2}\cdot \frac{1}{1-\frac{1}{2}}=\frac{1}{2}\cdot 2=1$$
So just before that n Xiang He $S_n=1$, It proves that this person can still walk this distance in the end , No less , Gratifying congratulations .