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In proving convergence of sequences using this definition:

A sequence $(π‘Ž_𝑛 )$ converges to $𝑐$ if for every $\epsilon>0$, there exists an index $𝑁$ so that, for all $π‘˜β‰₯𝑁$,

$|π‘Ž_π‘˜βˆ’π‘|<πœ–$.

if we pick really large epsilon, it is always going to work. Then what does the definition mean?

Another question on the same definition- as we know that the sequence $a_n=\frac{n^2}{n^2+1}$ is convergent but if we prove it using definition we find that for the sequence to be convergent we must have $0<\epsilon<1$ which means it doesn't work for every epsilon.

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  • $\begingroup$ The trick is, you want to pick a very small epsilon, and regardless of how small it is, I can also pick an N to satisfy the equation. The point is that its like a game. "I can pick a small enough epsilon you can't beat", "No, I beat you with my choice of N" $\endgroup$ – Don Thousand Sep 18 '18 at 15:55
  • $\begingroup$ But it doesn't say in the definition that the epsilon has to be small. It says " for every epsilon>0". $\endgroup$ – math Sep 18 '18 at 16:11
  • $\begingroup$ Yes, it means it should be true for every $\epsilon\gt 0$, be it small or large. $\endgroup$ – user418131 Sep 18 '18 at 16:34
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    $\begingroup$ Consider this - for two real numbers $x$ and $y$, I claim that if for every $\epsilon\gt 0, |x-y|<\epsilon$, then $x=y$. Do you see why this is true? $\endgroup$ – user418131 Sep 18 '18 at 16:43
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    $\begingroup$ To answer your question, we don't place restrictions on $\epsilon$. Suppose that $|a_n-1|\lt\epsilon$ for a particular $\epsilon\gt 0\ \forall n\ge\ N(\epsilon)$. Then $|a_n-1|\lt \epsilon'$ holds for all $\epsilon'\ge\epsilon$, for $n\ge N$ $\endgroup$ – user418131 Sep 18 '18 at 16:48
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Let us prove the convergence of $a_n$ to 1.

Claim : For all $\epsilon\gt 0\ \exists\ N(\epsilon) \in\mathbb{N}$ such that for all $n\in\mathbb{N}, n\ge N$,

$|a_n-1|<\epsilon$

Proof : Fix an $\epsilon \gt 0$

We note that $a_n$ is increasing, and $\forall n, a_n\lt 1$.

Therefore if for some $N(\epsilon),1-a_N\lt\epsilon$, then for all $n\ge N,1-a_n\lt\epsilon\;$ (We remove the modulus because $a_n\lt 1$)

Let us find such an $N$. $$1-a_N\lt\epsilon\Rightarrow \frac1{N^2+1}\lt\epsilon\Rightarrow N^2\gt\frac 1\epsilon-1$$

If $\frac 1\epsilon -1\lt 0$, then the inequality is true for all $N$. If $\frac 1\epsilon -1\ge 0$, then $N$ should be greater than $\sqrt{\frac 1\epsilon -1}$.

That's it. For an arbitrary $\epsilon\gt 0$, we found an $N(\epsilon) \in\mathbb{N}$ such that for all $n\in\mathbb{N}, n\ge N$,

$|a_n-1|<\epsilon$, and so $\lim_\limits{n\to\infty}a_n=1$

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  • $\begingroup$ is there an example where i would not be able to find N corresponding to an $\epsilon$? $\endgroup$ – math Sep 18 '18 at 17:48
  • $\begingroup$ Just consider this example itself. Show that, say, the limit of $a_n$ is not $\frac 12$ $\endgroup$ – user418131 Sep 18 '18 at 17:50
  • $\begingroup$ I have to find an $\epsilon>0$ for which there is no $N$ such that $\forall k\geq N$, we have $$\abs{a_k-\frac{1}{2}}<\epsilon$$. After simplifying, $$\abs{\frac{n^2-1}{2(n^2+1)}}<\epsilon \implies |n^2-1|<2\epsilon (n^2+1).$$ Since $n\geq 1$, we have $n^2-1\geq 0$. On simplifying, $$n^2(1-2\epsilon)<1+2\epsilon.$$ It gives rise to two cases: (1) $1-2\epsilon >0$ (2) $1-2\epsilon <0$. For (1), $$n^2<\frac{1+2\epsilon}{1-2\epsilon}$$ and because it gives an upper bound, this doesn't work. For (2), $$n^2<\frac{1+2\epsilon}{1-2\epsilon} $$ $\implies$ $N=\frac{1+2\epsilon}{1-2\epsilon}+1$. $\endgroup$ – math Sep 18 '18 at 18:14
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    $\begingroup$ Well, if you want to find a particular $\epsilon\gt 0$, fix one and show that the condition is not valid. In the above, take $\epsilon=0.1$ $\endgroup$ – user418131 Sep 18 '18 at 18:26
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    $\begingroup$ makes sense! Thanks a ton for your time! $\endgroup$ – math Sep 19 '18 at 15:46
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You have almost answered your own question, the challenge is to prove that it works for every possible $\epsilon$ that is positive.

Whenever an $\epsilon$ is chosen, you must be able to find the corresponding $N$ to show that it converges. If an $N$ works for a small $\epsilon$, it works for a bigger $\epsilon$ as well.

Edit:

You claim that if $0 < \epsilon < 1$, then if $N > \sqrt{\frac1{\epsilon}-1},$ then if $k > N$ we have $|a_k - c| < \epsilon$.

Using what you claim, we know that if $N> 1$, if $k > N$, then we have $|a_k-c| < \frac12$.

Now what if we are given $\epsilon$ that is at least $1$?

If $N>1$, then for any $N>1$, if $k > N$, we also have $|a_k - c| < \frac12 < 1 \le \epsilon$.

Also, actually, it is obvious that $\left| \frac{1}{k^2+1}\right| \le \frac12$ for all $k$.

Hence it is true that $\left| \frac{1}{k^2+1}\right| \le 1$ for all $k$, $\left| \frac{1}{k^2+1}\right| \le 2$ for all $k$ and so on.

An analogy to why focusing on small $\epsilon$ suffices is suppose from next month onwards, I will have less than $1000$ dollar for my salary, can I say that for sure I have less than $2000$?

Edit:

We want to prove that $\forall \epsilon > 0, \exists N_\epsilon>0, \forall n > N_\epsilon, |\frac{n^2}{n^2+1}-1| < \epsilon .$

That is we want to show that $\forall \epsilon > 0, \exists N_\epsilon>0, \forall n > N_\epsilon, \frac{1}{n^2+1} < \epsilon .$

We consider the cases where $\epsilon \ge 1$ and $0 < \epsilon <1$ separately.

If $\epsilon \ge 1$, we pick $N_\epsilon=1$, then $n> N_\epsilon=1$ means $n^2 > 1$, $n^2+1 > 2$ and hence $\frac1{n^2+1}< \frac12 < 1 \le \epsilon.$ We have considered the case when $\epsilon \ge 1$.

If $0 < \epsilon <1$, we pick $N_\epsilon = \sqrt{\frac1{\epsilon}-1}$, hence $n > \sqrt{\frac1{\epsilon}-1}$ implies $n^2 +1 > \frac1{\epsilon}$, and we have $\frac1{n^2+1} < \epsilon.$

We have considered every positive $\epsilon$ and find a corresponding $N_\epsilon$ such that the condition holds. Hence it converges.

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  • $\begingroup$ does that mean the sequence n^2/(n^2+1) does not converge? $\endgroup$ – math Sep 18 '18 at 15:58
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    $\begingroup$ It converges. sometimes when you read a proof and you see that people focus on smaller values of $\epsilon$, for example here, you see $0 < \epsilon <1$, the thing is suppose it works for $\epsilon$ that is small, if you ever have to work with a bigger number, say $\epsilon_2$, if we can find an $N$ such that $|a_n -k|<\epsilon$ then clearly for the same $N$, we have $|a_n -k|<\epsilon < \epsilon_2$. It is not that it doesn't hold for bigger $\epsilon$, bigger $\epsilon$ is not the challenge and it is known that it holds once we deal with smaller $\epsilon$. $\endgroup$ – Siong Thye Goh Sep 18 '18 at 16:04
  • $\begingroup$ I am not able to understand your argument. To help stating the problem clearly: $$|a_k-c|=|\frac{π‘˜^2}{π‘˜^2+1}-1|=|\frac{k^2βˆ’(k^2+1)}{k^2+1}|=|\frac{1}{k^2+1}|<\epsilon$$ Then, solving for $k$ in terms of $\epslion$ yields a formula : |1/(π‘˜^2+1)|=1/(π‘˜^2+1)<πœ–βŸΉ1/πœ–<π‘˜^2+1 β‡’βˆš(1/πœ–βˆ’1)<π‘˜ So, as long as πœ–<1, we can pick N to be the first positive integer larger than √(1/πœ–βˆ’1). $\endgroup$ – math Sep 18 '18 at 16:15
  • $\begingroup$ see the edit and see if it helps? $\endgroup$ – Siong Thye Goh Sep 18 '18 at 16:23
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    $\begingroup$ see edit. focusing on small $\epsilon$ actually suffices but perhaps for now just consider every $\epsilon$ explicitly if it's more comfortable for you. $\endgroup$ – Siong Thye Goh Sep 18 '18 at 17:08

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