convergence of a sequence- choosing epsilon and understanding

Question

In proving convergence of sequences using this definition:

A sequence $(_ )$ converges to $$ if for every $\epsilon>0$, there exists an index $$ so that, for all $≥$,

$|_−|<$.

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.

Answer 1

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 $\lim_\limits{n\to\infty}a_n=1$

Answer 2

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.