How to judge whether a sequence has a limit

How to judge whether a sequence has a limit

1. Conceptual method: there is a positive number ε, When n > N, | an-m |< ε equation holds good under all circumstances
2. Theorem method:
(1) Monotone and bounded sequence must have limit;
(2) Pinch criterion;
(3) Mathematical induction (may be used in combination with (1) and (2))
3. Function method: form the general term formula of the sequence into a function, and determine the limit of the sequence by finding the limit of the function, which should be used together with the pinch criterion or conceptual method
1. Prove that the limit of sequence {xn = (n-1) / (n + 1)} exists and find its limit
prove:
∵1 -1/(1+1/n) = 1- n/(n+1)< 1-2/(n+1) = xn < (n-1)/n = 1-1/n
Namely: 1 - 1 / (1 + 1 / N) < xn < (n-1) / N = 1-1 / n
Known: when n is infinite: Lim 1 / N = 0
∴lim[1 -1/(1+1/n)]=1
lim[1-1/n]=1
According to the clamping force alignment side: xn limit exists and limxn = 1
2. Omitted, the method is the same as 1

The problem of judging right and wrong by sequence limit 1. In the definition of sequence limit ε Is an arbitrarily small positive number 2. There are infinite numbers of N in the limit of sequence, but it is enough to find one 3. If a sequence has a limit, then the limit is unique 4. And | an-a|

1. In the definition of sequence limit ε Is an arbitrarily small positive number
[answer] Yes. Only when it can be arbitrarily small can it explain the infinite approach, that is, the existence of limit
2. There are infinite numbers of N in the limit of sequence, but it is enough to find one
[answer] Yes. As long as n is greater than N, the inequality holds. There are countless numbers greater than N, which can be used as n
3. If a sequence has a limit, then the limit is unique
Yes. Even if it is fluctuating, it is not a limit, but only bounded
4. And | an-a|

The method of judging the existence of sequence limit based on sequence limit A method for judging the existence of sequence limit

If you tell a recursive formula, the general method is the monotone bounded method. As long as you prove that its monotone increase has an upper bound or monotone decrease has a lower bound, it shows that the limit of the sequence exists. How much is it, that is, take the limit on both sides of the recursive formula (you can also use the definition, which is when there is no monotonicity, that is, you first find the limit on both sides of the recursive formula to obtain the limit value of the sequence. This process is on the draft paper, and then use xn or xn + 1 to subtract the value you find to obtain the absolute value. Use the recursive formula to continuously enlarge the formula so that it is less than. And finally less than a formula related to N, and when n tends to infinity If the formula is equal to 0, it means that the limit of the sequence exists and is equal to the value you start to seek. If you don't understand this method, it's OK. Generally, it won't be tested.)
If what you are given directly is the general term of the sequence, then you can directly find the limit of the general term to see if it exists

How to judge whether a sequence is divergent or convergent, and how to find the limit of a sequence

When n tends to infinity, it tends to a certain value, which is convergence, otherwise it is divergence
Your second question is so good that you can write half a book

Proof of limit of simple sequence of Higher Mathematics According to the definition of sequence limit, it is proved that LIM (1 / N ^ 2) = 0 (n goes to 1 to positive infinity) If I can do it, I don't have to send it out

Do it yourself. It's so simple

Proof of sequence limit Certificate: if (1)y(n+1)>y(n) (2)lim yn->∞ (3) LIM (x (n + 1) - x (n)) / (Y (n + 1) - yn) exists Then Lim xn / yn = LIM (x (n + 1) - x (n)) / (Y (n + 1) - yn)

Still the same, the definition of limit, infinite, finite + infinite
LIM (x (n + 1) - x (n)) / (Y (n + 1) - yn) exists
Let LIM (x (n + 1) - x (n)) / (Y (n + 1) - yn) = a
For any E > 0, there is n such that there is a difference for n > n
|(x (n + 1) - x (n)) / (Y (n + 1) - yn) - a| then for n > N, there is
A-E < (x (n + 1) - x (n)) / (Y (n + 1) - yn) (A-E) (Y (n + 1) - yn) then
(a-e)(y(N+2)-y(N+1))(a-e)(y(N+3)-y(N+2))...
The addition of (A-E) (Y (n + 1) - yn) has
(A-E) (Y (n + 1) - Y (n + 1))
|(x (n + 1) - x (n + 1)) / (Y (n + 1) - Y (n + 1)) - a | is now transformed into xn / yn containing the above formula and its limit is proved
Xn / yn - a = (xn-x (n + 1)) / (yn-y (n + 1)) * (yn-y (n + 1)) / yn + (x (n + 1) - A * y (n + 1)) / yn find out the above formula
|xn/yn - a|<=e|1-y(N+1)/yn|+|(x(N+1)-a*y(N+1))/yn|<=e+|(x(N+1)-a*y(N+1))/yn|
The existence of n '> n makes it possible to have an effect on N > n'
|(x (n + 1) - A * y (n + 1)) / yn| then for any E > 0
Yes | xn / yn - a | < 2e
Then Lim xn / yn = LIM (x (n + 1) - x (n)) / (Y (n + 1) - yn)