Finding the Limit of the Number Sequence Mathematics Mathematical Problems Mathematical Analysis Fin

Don't use numbers to prove it, or you'll fall into a trap. Obviously, it is converted into integrals, and it depends on whether you understand the essence of riemann integrals. If you try to do it with the limit of the number series, you may never be able to do it.

Original = (1 n)*

The integrand is f(x)=1 (4-x 2), and the integral interval is 0 1, divided into n parts, each of which is 1 n in length.

When n +, the original formula = [1 (4-x 2)]dx, which is accumulated from 0 to 1.

Obviously, this integral exists, and the value of the limit of the original problem is the value of this integral.

Let x=2sin, and the corresponding integral interval becomes 0 6, and the integral formula has [1 (4-x 2)]dx= [1 (4-4(sin) 2)]d(2sin)= d = 6-0= 6

So, the limit of the original sequence exists, which is 6.

Integrals to solve, answer 6

See references for details.

First of all, I give 3 conclusions without proof:

If it converges at the same limit a, then converges at a

If the sequence converges to a, then any of its subcolumns also converges to a

If a series of numbers converges, then its limit is unique.

Secondly, because the condition of the question is that and both converge. Suppose that this sequence converges at a, and if the sum can be deduced and also converges at a, then the answer comes out.

Consider a sequence, which is both a subcolumn and a subcolumn. Now due to convergence at a, according to conclusion 2, convergence at aIt is then assumed that convergence is at b, and according to conclusion 2, convergence at bThen according to conclusion 3, a=b, that is, converges to a

Considering the sequence again, it is a subcolumn of and in the same way converges to a

So according to Conclusion 1, convergence.

Zhang Yu's information on this type of question is summarized, and the method is cut first and then played. This means that if you find the relationship between xn-a and x(n-1)-a, and then iterate over x1-a, you will find that the previous coefficient is 0 to the nth power, and the force converges.

The limit is zero, and when n approaches infinity, n is much greater than n, so n can be ignored, and the result is a constant divided by infinity, which is equal to 0

Or you divide the original formula into three parts, find the limits separately and add them together, and the limits of each part are zero, so the result is zero.

For the sake of convenience, I will write down the same formula you have above as sn

Obviously n (n2+n+n) So, the limit you are looking for is 0

n/(n^2+n+n)《

1/(n^2+n+1)+1/(n^2+n+2)+.1/(n^2+n+n)

n/(n^2+n+1)

Since limn (n 2+n+n)=limn (n 2+n+1)=0 is affected by the pinch theorem: the original limit is 0

Solution: (1) From, xn+1= 1 2(xn+a xn), and x1>0 know xn>0, xn+1= 1 2(xn+a xn) a, xn= a, (in fact, if x1 = a, then the whole column xn is a) The following discussion excludes x1 = a:

xn+1) (xn) 1 2(1 a x n) 1, i.e., xn+1 xn, so the series is monotonically reduced and has a lower bound.

2) Let lim xn (n )m, and take the limit on both sides of xn+1 = 1 2(xn+a xn) at the same time, and obtain:

m=1 2(m+a m), solution; m＝﹙1＋√(1+4a)﹚/2

xn=[3^n(1/3^n+(2/3)^n+1]^(1/n)

3*[1/3^n+(2/3)^n+1]^(1/n)

When n is hidden, 1 3 n 0, (2 3) n and hail 0, the sail is called, xn 3*1 0=3

You don't understand the definition problem; The limit is that if you give any >0, you can find an n, so that when n takes n > n, there is |xn-a|<ε

The key is whether we can find this n, and the method you wrote is to work backwards and use |xn-a|< Push the range of n and then push n

Specific: Arbitrarily give a , 0 (let <1), as long as 1 (n+1)< according to |xn-a|=|(-1)^n/(n+1)^2-0|=1 (n+1) 2<1 (n+1)< Derives |xn-a|< Apparent inequality |xn-a|< must be true.

1 (n+1)< gives n>1 -1, which can take n=[1 -1], when n>n>=1 -1, i.e. 1 (n+1)< then the inequality |xn-a|< must be true.

I don't understand and ask.

Inequalities|xn-a|< must be true. The definition of the limit requires that this inequality exist.

xn-a|It is the distance between the terms of the series and a. This distance requires "as small as it can be".

It's as small as it gets. Therefore, this inequality must be true in order for us to be justified in saying that the limit of xn is a. The question is, when does this inequality hold?

This requires a solutionxn－a|< This is an inequality.

For the purposes of this question: |xn－a|=1 (n+1) 2, so this inequality is equivalent to solving.

1/(n+1)^2<ε。This inequality can already be solved, n+1>1 root number ( ) or n>1 root number ( )1. That's already okay.

But a lot of questions|xn－a|< This inequality is difficult to solve.

So we can take a scaling approach. Like this question, I know that 1 (n+1) 2<1 (n+1), if 1 (n+1)< is true, then.

xn－a|=1 (n+1) 2<1 (n+1)< is true, and inequality 1 (n+1)< is easy to solve. Therefore, many of the problems that use the definition and follow are scaling techniques that use inequalities.

The first one uses Robida's rule to examine the knowledge of variable upper limit integrals. (In the denominator, the integral number is 2+x, not 2+t??.)）

The second mecca theorem, first the terms are 0, so the limit is non-negative, and then each term is the first term, there are n terms, so that the limit of the first term *n = 0. The limit is 0

1.Apparently 0 0

Lobida. [e^(x^2)*(x^2)'-0]/[1)∫[0,x]e^(2t^2)dt+(2+x)e^(2x^2)]

2xe^(x^2)/[1)∫[0,x]e^(2t^2)dt+(2+x)e^(2x^2)]

Or 0 0 again Lopida.

2e (x 2)+4x 2e (x 2)] e (2x 2)+e (2x 2)+(2+x)*4xe (2x 2)] to x=0

2.=lim n->infinity 1 (n 2+n+1)+lim n->infinity 1 (n 2+n+2)+lim n->infinity 1 (n 2+n+n).