Question

Please help! Only answer questions 5-8!

Definition 0.1. A sequence X = (xn) in R is said to converge to x E R, or x is said to be a limit of (xif for every e > 0 there exists a natural number Ke N such that for all n > K, the terms Tn satisfy x,n - x| < e. If a sequence has a limit, we say that the sequence is convergent; if it has no limit, we say that the sequence is divergent Check to see if each of the following statements is equivalent to the definition of limit. If it is, prove it. If it is not, give an example to justify your conclusion 1. For every e > 0 there exists a natural number Ke N such that for all n > K, the terms x,n satisfy nx| < e. 2. For every e >0 there exists a natural number Ke N such that for all n > K, the terms xn satisfy |xn - x| < 100e 3. For every e >0 there exists a real number Ke R such that for all n > K, the terms xn satisfy 4. For every e > 0 there exists a natural number Ke N such that for all n > K, the terms xn satisfy rn | < €e. 5. For every rational number e > 0 there exists a natural number Ke N such that for all > K the terms xn satisfy |xn - x| < €. 6. For every e > 0, for every natural number Ke N, there exists n E N such that n > K and the terms an satisfy |xn - x| < e. 7. There exists a number e > 0, there exists a natural number Ke N such that for all n > K, the terms xn satisfy |xn - x| < €. 8. For everye > 0, for all n e N, the terms xn satisfy xn - x < e.

Answer 1

5) This statement is equivalent as for any real number
L
, there exists rational $\epsilon'>0$ such that
EE
, and then use the definition.

6) Note that this is not true consider the sequence . As then by this definition this sequence converges to both 1 and -1, which is a contradiction as the sequence does not converges by the original definition.

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7) This is also not equivalent. Consider the same sequence. Then for all $n\ge 1$, , but the sequence is not convergent .

(-1)"-0 2

8) That not equivalent. Note that this condition is a stronger condition (unlikely the other two) that is this statement implies the sequence is convergent but not conversely (by this definition only constant sequence is convergent). Consider the sequence $\{\frac{1}{n}\}$ . Then this converges to 0. But note that choose $\epsilon>\frac{1}{2}$ , then
.이f€
, for
n =
.

Feel free to comment if you have any doubts. Cheers!