Why other definitions of convergence fail to be correct? 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 $|a_n - a| &lt; \epsilon$ for all indices $n \ge N$. 


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

Please help!
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 $|1/n-1| &lt; \epsilon$ for every $n&gt;N$.

So it is true that for some $\epsilon&gt;0$ there is $N$ so $|1/n-1|&lt;\epsilon$ for every $n&gt;N$. 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 $\epsilon&gt;0$ we have $|1/n-0|&lt;\epsilon$ for every $n&gt;N$. Given $N$, let $\epsilon=N/3$ and $n=2N$. Then $n&gt;N$ but $|1/n-0|&gt;\epsilon$.
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?