F (x) is monotonically bounded in the positive and negative infinite interval, and xn is a sequence. If xn is monotonic, then f (x) converges? If f (x) = 1 / X and xn = 1 / N, this conclusion is not tenable

F (x) is monotonically bounded in the positive and negative infinite interval, and xn is a sequence. If xn is monotonic, then f (x) converges? If f (x) = 1 / X and xn = 1 / N, this conclusion is not tenable

There is something wrong with your description of the problem. It should be f (xn) convergence. This is for sure
Your example is incorrect
Note that f (x) = 1 / X is monotonic but not bounded
The convergence of F (xn) is proved
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If xn is monotone and bounded, then it must have a limit, set as a
Then f (xn) tends to f (a), so it converges
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If xn is monotonous but unbounded, then it tends to ∞,
Note that if f (x) is monotonically bounded, there must be a limit,
That is, when x is sufficiently large, f (x) has a limit and is set to B
Slightly more strictly, the language is described as:
There is a positive number m, so that when | x | > m, LIM (x →∞) [f (x) - b] = 0
in other words,
When x →∞, LIM (x →∞) [f (x) - b] = 0
Then f (x) naturally converges
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The certificate is completed
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Function f is bounded on R and xn is a sequence. What is the relationship between xn and f (xn)?

Xn is a sequence, so f (xn) is also a sequence, and f (xn) is a sequence obtained by xn according to the rule F. since f is specified to be bounded on R, even if xn is unbounded, the obtained sequence f (xn) is bounded

Can "convergence must be bounded" in sequence be extended to functions

Yes, but there are differences
The function is locally bounded and the sequence is globally bounded
If limf (x) = a (x tends to infinity), then there is x > 0. When | x | > x, f (x) is bounded
If limf (x) = a (x tends to x0), there is a > 0 when x0-a

Convergence function and subsequence problem For the sequence {xn}, if x2k-1 approaches a (k approaches positive infinity) and X2K - approaches a (k approaches positive infinity), it is proved that xn approaches a (n approaches positive infinity)

Proof 1: use Cauchy convergence theorem. That is, when k is infinite, any two terms can be infinitely close. Here, a can be an excessive intermediate quantity. First let the odd term be half of epsilon, and even numbers, and then combine it with absolute value inequality. Proof 2: directly use the limit theorem. When k is infinite, odd numbers

It is proved that the function series (- 1) ^ n / (x + 2 ^ n) converges uniformly at (- 2, positive infinity) Can you use the m-discriminant method? Urgent

The first term can be removed, and then the control series can take (- 1) ^ n / (2 ^ n-2), or directly use the Dirichlet discriminant method

Let function f (x)= Σ (x + 1 / N) ^ n, (1) find f (x) definition field D (2) prove that the series does not converge uniformly on D

1 it is known from the root value discrimination that the convergence domain is R 2 by the Cauchy convergence criterion (the upper and lower serial numbers are limited A1 and A2), then x can always be taken as sufficient to make it greater than ε Then inconsistent convergence