Why other definitions of convergence fail to be correct?

Question

The following is an exercise from the book Advanced Calculus:

Well obviously the b. is not correct since for $\epsilon= \frac19$, ${\{a_n}\}={\{\frac1n}\}$ and $N=2$ it fails to be correct. But the a. and the c. are exactly as same as the accepted definition of convergence which is given in the book, i.e.

Definition. A sequence ${\{a_n}\}$ is said to converge to the number a provided that for every positive number $\epsilon$ there is an index $N$ such that $|a_n - a| < \epsilon$ for all indices $n \ge N$.

In fact, the a. and the c. are just re-phrasing the above mentioned definition.

Please help!

Answer 1

They are not the same! For the first, "For some" means "for at least one but not necessarily all". Here, you want to use that if you can't choose an arbitrary $ε$, then $1/n$ will converge to more than one number.

For the third, "For all A there exists B" is very different from "there exists B for all A". For every cow there is a glass of milk vs for every glass of milk there is a cow. if $1/n$ satisfied this definition it would be identically zero for $n>N$.

Answer 2

For $a$, the problem is the phrase for all indicies. Choose small number $\epsilon$ and go to work. For $c$, the problem is the phrase There is an index N. If you fix $N$ in advance and then choose epsilon later, you will run into a problem. So given $N$, how do you choose $\epsilon$ to make your counterexample $a_n = \frac{1}{n}$ work?

Answer 3

No, a and c are not the same! Those changes you call "rephrasings" totally change the meaning - that's the point to the exercise.

Let $\epsilon=3$. Let $N=1$. Then $|1/n-1| < \epsilon$ for every $n>N$.

So it is true that for some $\epsilon>0$ there is $N$ so $|1/n-1|<\epsilon$ for every $n>N$. Which is what a says - if a is the definition of limit we've just proved that $$\lim_{n\to\infty}\frac1n=1.$$

And if c were the same as the actual definition that it would not be true that $\lim 1/n = 0$. Because c does not hold here - it is not true that there is an $N$ such that for every $\epsilon>0$ we have $|1/n-0|<\epsilon$ for every $n>N$. Given $N$, let $\epsilon=N/3$ and $n=2N$. Then $n>N$ but $|1/n-0|>\epsilon$.