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Definition: A sequence $(a_n)$ converges to a real number $a$ if, for every positive number $\epsilon$ , there exists an $N \in\mathbb{ N}$ such that whenever $n \ge N$ it follows that $|a_n − a| <\epsilon$.

Example: Find $\lim_{n\to\infty}\frac 1n$.

Using the def.: Let $\epsilon> 0$ be arbitrary. Choose $N\in\mathbb{N}$ with $N > 1/\epsilon$. To verify that this choice of $N$ is appropriate, let $n\in\mathbb{N}$ satisfy $n\ge N$. Then, $n ≥ N$ implies $n > 1/\epsilon$, which is the same as saying $1/n < \epsilon$. Finally, this means $|1/n-0|<\epsilon$ as desired.

But here I guessed the value of limit point $a$ (using calculus) around which I devoted my neighborhood. What if I took $a=1$ instead?

Let $\epsilon> 0$ be arbitrary. Choose $N\in\mathbb{N}$ with $N > \frac{1}{\epsilon+1}$. To verify that this choice of $N$ is appropriate, let $n\in\mathbb{N}$ satisfy $n\ge N$. Then, $n ≥ N$ implies $n > \frac{1}{\epsilon+1}$, which is the same as saying $1/n < \epsilon+1$. Finally, this means $|1/n-1|<\epsilon$.

Where am I mistaken?

Please help I am new in Real Analysis, thank you.

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The very last step is wrong. From $1/n<\epsilon+1$ you can get $1/n-1<\epsilon$, but not $|1/n-1|<\epsilon$. Indeed, since $1/n<1$ for $n>1$, $1/n-1$ will be negative, so $|1/n-1|=1-1/n$, which will probably be larger than $\epsilon$.

This wasn't an issue in the first proof because $1/n-0=1/n$ is positive, so taking the absolute value doesn't change it.

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  • $\begingroup$ I see there can't be a point greater than the limit in general, and cant be smaller too, as then our choice of $\epsilon$ will be limited. Is this correct? $\endgroup$
    – gambler101
    Sep 4 '16 at 23:26
  • $\begingroup$ I'm not sure what you mean by that. It is certainly possible for there to be values of $n$ such that $a_n>a$ and also different values of $n$ such that $a_n<a$. $\endgroup$ Sep 4 '16 at 23:46
  • $\begingroup$ I meant that if $a_n>a$ than the choice of $\epsilon$ is not arbitrary? $\endgroup$
    – gambler101
    Sep 4 '16 at 23:50
  • $\begingroup$ I still don't know what you mean. It's OK if $a_n>a$, since you get to chose $N$ based on $\epsilon$, and so if there are some $n$ such that $a_n>a+\epsilon$ you may be able to choose $N$ large enough so that no such $n$ is greater than $N$. $\endgroup$ Sep 4 '16 at 23:55
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    $\begingroup$ Well, $1/n-a$ will be positive then, but you won't be able to prove that $1/n-a<\epsilon$ to begin with. $\endgroup$ Sep 5 '16 at 0:01

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