When n approaches infinity, the power of n and n log is based on 2 and n is logarithmic is larger

The result is n!Big.

Tannd didn't sleep to do your questions, and when you finished you found what you had done before.

This is a mathematical analysis question.

First of all, take the logarithm of the two numbers to be compared, (for e), although in the end it is when infinity is considered, but this makes a pattern.

After taking the logarithm separately, the first one is equal to a=ln1+ln2+..ln(n) the second is equal to b = log2(n) * ln(n) = ln2**a by plotting him in the coordinate system, we can know his relation to the integral of the function lnx:

a> (1,n)ln(x)dx =c. (There's a simple proof here, so I won't go into it for the sake of length.) The monotonicity of the function ln(x) is used to make use of the non-negativity of ln(x) on the interval (1,n) and the definition of the integral is easy to prove) (1,n) indicates that the lower bound of the integral is 1 and the upper limit is n

Using calculus to calculate c=n*ln(n)-n+1, calculate g(n)=c-b=n*ln(n)-n+1- ln2 To avoid the following troubles, count 1 ln2=k

Derivative, g'(n)=ln(n)-2k*ln(n) n when n tends to infinity, ln(n) n tends to 0, and ln(n) tends to infinity, so g'(n) tends to infinity. So g(n) also tends to infinity.

This proves that when n tends to infinity, a>b

At the same time, the two numbers that were originally compared, E a and E b, are monotonicity known when n tends to infinity! Bigger!

Please check!

geniu007：

How to prove lg2 (lg10 n!) by derivative- lgn) greater than 0?

What is puzzling is that n (log is based on 2 and n is logarithmic) to the power and n!are all constants, so lg2 (lg10 n!- lgn) is a constant.

There is only one result of finding the derivative of a constant, and that is equal to 0.

Don't forget the first sentence of the question, "When n tends to infinity".

Why was the level so low in previous years? Didn't you study math well?

Everyone has to study and be annoyed often, and the questions should not be so excited.

Younger students of this year! I don't even know how to study math well! Such a simple question is really not the level of the college entrance examination, the level of graduation is high!

Okay, I'll do it!

n！-log is based on 2 and n is logarithmic.

The result can be written as n!-lgn lg2 (the formula for swapping the base) is further equal to lg2 (n!-lgn）

lg2 （lg10~n!- lgn）

Next, find the derivative pair (lg10 n!- LGN).

Sorry, I haven't read a book in a year! How to find the derivative to forget the light! Oh, find the derivative function to prove that it is greater than 0, which is n! Big.

The limit to the nth power is 1 e,This makes use of an important limit = [1-1 (n+1)] n+1)*(n) (n+1)]; =e^(-1)。When n->, lim (1+1 n) n=e.

Therefore, lim (n (n+1)) n=lim 1 (1+1 n) n=1 e mainly uses the trick of n=1 (1 n), so n (n+1)=1 (n+1) n)=1 (1+1 n).

The equation of infinite symbols

In mathematics, there are two equations of infinite notation that are occasionally used, namely: = +1, = 1.

A positive value represents an infinite number of formulas, without a specific number, but a positive infinity means a value that is greater than any one number. The symbol is +, and the symbol for negative infinity is -.

The Möbius strip is often considered the idea of the infinity symbol "", because if someone stands on the surface of a giant Möbius strip and walks all the way down the "road" that he can see, he will never stop. But this is an untrue rumor, because the invention of " " was earlier than the Möbius strip.

First write it as (-(n+1)(-n) (n+1)) (n+1)) of (1-1 n+1), then let n be equal to 1/x of x, and solve the negative of e according to the 1 x power of that lim(x tends to 0) (1-x) is equal to e.

When it is 1 to the infinity power.

limu^v=e^lim(u-1)v

Original = e -1

（n/n+1）^n=(1-1/n+1)^n=[(1-1/n+1)^-n+1)]*1-1/n+1)

According to the important limit formula, when x->, lim(1+1 x) x=e, so that x=-(n+1), so the original limit.

lim（n/n+1）^n=lim[(1-1/n+1)^-n+1)]*1-1/n+1)=e （n->∞

When n tends to infinity, n approximate equals n+1, so n n+1 approximate equals 1, then the infinity square of 1 is still equal to 1

Calculated with special limits as follows, n (n+1)) n = lim (1-1 (n+1)) n = lim (1-1 n) (n)*(1) = e (-1).

Calculated with special limits as follows, click to enlarge:

Note a=(2n+1)!/2n)!=1/2)*(3/4)*.2n+1) and sell 2n

Then 00 (n tends to be the end of the hole, teasing and seeping infinite).

n! >2 n > 10n 2 > 100n > 15n+100log n>log n 3 > log n e 10

Taking n as the variable, the following is sorted from fast to slow when tending to infinity.

n to the n power, n to the deliberative power, a to the n power (exponential function) a>1, n to the a power (power function) a>0, logarithmic function ln(n).

Several common functions that tend to infinity can be in this order, and if you encounter them when doing the problem, you can directly compare the magnitude to get the result.

For example, x tends to positive infinity x e x, the direct result is 0, x tends to 0+, xlnx can directly make the positive limb result 0, and so on.

Extended information: The growth curve model as a whole presents an "S" shape, which can be divided into three stages: the early stage, the middle liquid stage, and the end stage

1) In the early stage, although x is in the growth stage, y grows slowly, and the curve shows a relatively gradual upward trend.

2) In the medium term, with the growth of x, the growth rate of y gradually increases, and the curve shows a rapid upward trend;

3) When the inflection point (x*,y*) is reached, the growth of the saturation degree of the function reaches the end, and with the growth of x, y grows slowly, and the growth rate tends to be close to 0, and the curve develops horizontally.

Teach you an important limit of stupidity.

For (1+1 n) n

n--> infinite.

1+1 n) n = e lim(1 n)*n, that is, Chi Xin said lim (1 + infinity about n) Infinity about n = e lim (infinitesimal about n * infinity about n).

(n+1)=e lim(2 (2n+1))*n+1)=e (1 2).