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${\displaystyle (a_n)}$ is a sequence with ${\displaystyle a_n = \frac{1}{\sqrt{n}}}$. Proove that ${ \displaystyle \lim\limits_{n\to\infty}{a_n} = 0 }$, using epsilon-delta method.

First of all, I assume that ${ \displaystyle n \in [1;+\infty) }$.

For all ${n}$, which are larger or equal to ${1}$, ${n^{th}}$ term of sequence is strictly larger than zero and less or equal to one.

${ \forall n \geq 1 : 0 < a_n \leq 1 }$

Assume some ${ \displaystyle \epsilon > 0 }$, such that ${ \displaystyle 1 - \frac{1}{\sqrt{n}} < \epsilon \Longleftrightarrow \left(\frac{1}{1-\epsilon}\right)^2 > n }$

Now let ${ \displaystyle N = 1 + \left\lceil \left(\frac{1}{1-\epsilon}\right)^2 \right\rceil }$, then ${ \displaystyle \left| \frac{1}{\sqrt{n}} - 1 \right| < \epsilon }$ ${ \displaystyle \forall n \geq N }$

Have I understood the task? Is my solution correct?

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No. When proving a sequence converges to a limit, the phrase "Assume some $\epsilon>0$, such that ..." should NEVER appear. You assume $\epsilon>0$ is arbitrary and fixed, but cannot make any additional assumptions otherwise it isn't arbitrary. Second, if you show that $|a_n-1|<\epsilon$ then you are showing the sequence converges to $1$ and not $0$. – Matt Nov 24 '12 at 22:17
Another problem with your argument (which might go away if you address other problems) is that you conclude in line 4 that as long as $n$ is less than the given expression in $epsilon$, then you get your desired inequality. Then in line 5 you make a claim that applies to $n$ greater than that expression. – alex.jordan Nov 24 '12 at 22:18
Here is a related problem. – Mhenni Benghorbal Nov 24 '12 at 22:21
up vote 3 down vote accepted

As mentioned in comments, you should be investigating the inequality $\frac{1}{\sqrt{n}}<\epsilon$, not $1-\frac{1}{\sqrt{n}}<\epsilon$.

It's OK to do preliminary side work that establishes $$\frac{1}{\sqrt{n}}<\epsilon\iff n>\frac{1}{\epsilon^2}$$ and then use that to decide that a good choice for $N$ is $\lceil\frac{1}{\epsilon^2}\rceil$. But the logic of an epsilon-N argument doesn't start until:

  1. Let $\epsilon$ be someone's arbitrary idea of what "small" is. That is, let $\epsilon$ be some positive number that is handed to me by someone else.
  2. Turn around back to them and say "based on your $\epsilon$, here is an $N$-value: $\lceil\frac{1}{\epsilon^2}\rceil$. And I can demonstrate that '$n\ge N\implies \frac{1}{\sqrt{n}}<\epsilon$'."

That last directional implication must be in that direction for the logic of this argument to work.

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Thank you! One elaboration: is ${\lceil\frac{1}{\epsilon^2}\rceil}$ means integer part without rounding it or it is the next integer, which is larger or equal to ${\frac{1}{\epsilon^2}}$? – Edward Ruchevits Nov 24 '12 at 22:43
$\lceil x\rceil$ is the smallest integer that is greater than or equal to $x$. If $x$ is positive, it unambiguously means leave integers alone, and round up nonintegers. If $x$ is negative, you still round up, but keep in mind that makes absolute value be smaller. Similarly, $\lfloor x\rfloor$ works in the other direction. – alex.jordan Nov 24 '12 at 22:46

Assume some $ϵ>0$, such that $1−\frac1{\sqrt n}<ϵ$ […]

This hints at a serious misunderstanding. You appear to be assuming what it is that you have to prove.

"The enemy" comes with a devilishly clever choice of $\epsilon$ and you have to show that you can find an $N$ that demonstrates that "the enemy's" clever choice isn't good enough.

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Thank you, I hope I finally figured it out with the priceless help of SE mathematicians. :) – Edward Ruchevits Nov 24 '12 at 22:46

There are a few errors in what you have written. First you want to prove that the limit is $0$. In which case, you want to prove the following.

Given any $\epsilon >0$, there exists a $N(\epsilon) \in \mathbb{N}$, $$\text{ such that for all $n > N(\epsilon)$, we have }\left \vert a_n - 0 \right \vert < \epsilon \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$

When you are proving a limit from first principles, you cannot choose $\epsilon$; $\epsilon$ is given to you. The challenge for you is to find $N(\epsilon)$. So whatever be the '$\epsilon$' I give you, you should be able to find a $N(\epsilon)$ such that $(\star)$ is satisfied.

In your problem, given $\epsilon > 0$, you need to find $N(\epsilon) \in \mathbb{N}$, $$\text{ such that for all $n > N(\epsilon)$, we have }\left \vert \dfrac1{\sqrt{n}} - 0 \right \vert < \epsilon \text{ i.e. } \dfrac1{\sqrt{n}} < \epsilon $$ Now note that $a_n = \dfrac1{\sqrt{n}}$ is a decreasing sequence and hence if we find $m$ satisfying $(\star)$, for any $n > m$, $(\star)$ will be satisfied since for $n>m$, we have $a_n < a_m$. This motivates us to choose $N(\epsilon)$ as $$N(\epsilon) = \left \lceil\dfrac1{\epsilon^2} \right \rceil$$ Note that $$N(\epsilon) = \left \lceil\dfrac1{\epsilon^2} \right \rceil \geq \dfrac1{\epsilon^2} \implies \sqrt{N} \geq \dfrac1{\epsilon} \implies \dfrac1{\sqrt{N}} \leq \epsilon$$ Hence, for all $n > N$, we have that $$\dfrac1{\sqrt{n}} < \dfrac1{\sqrt{N}} \leq \epsilon$$

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Thank you for such a consistent and complete answer! – Edward Ruchevits Nov 24 '12 at 22:40

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