7
$\begingroup$

$$a_n=\frac{n^2+2n+6}{n^3-3}$$

So I want to show that "$a_n\to a\iff\forall \epsilon>0,\quad\exists N\in\mathbb{N}:n\geq N\implies |a_n-a|<\epsilon$"

Then my rough working:

$|a_n-0| =\left|\frac{n^2+2n+6}{n^3-3}\right|<\epsilon$

Estimate $\frac{n^2+2n+6}{n^3-3}<\frac{b_n}{c_n}$

Require $b_n>n^2+2n+6$, so choose $b_n=2n^2>n^2+2n+6$ for $n\geq 4$

Require $c_n<n^3-3$, so choose $c_n=\frac{n^3}{2}<n^3-3$ for $n\geq 2$

Then let $N=\frac{2n^3}{n^3/2}=\frac{4}{n}$

So now the proof:

$$\text{Fix}\quad \epsilon>0.\quad\text{Pick}\quad N:N>\frac{4}{n}\quad\text{and}\quad N\geq 4$$

$$\implies \forall n\geq N,\quad |a_n-0|=\left|\frac{n^2+2n+6}{n^3-3}\right|<\frac{4}{n}\leq\frac{4}{N}\leq\epsilon$$

Is this correct? Is it completely rigorous? I've just started doing analysis on my own, so please point out if anything is wrong/not conventional so that I can get into good habits at the start. Thanks for any help.

$\endgroup$
5
  • $\begingroup$ $N$ can't be related to $n$. You need to express it with the arbitrary $\epsilon$ $\endgroup$
    – user76568
    Commented May 10, 2017 at 20:23
  • $\begingroup$ @Dror, I see, so if I instead write "Pick $N>\frac{4}{\epsilon}$", does my argument hold? $\endgroup$
    – mrnovice
    Commented May 10, 2017 at 20:26
  • $\begingroup$ $N$ must be fixed "in response" to $\epsilon$ being arbitrary and fixed $\endgroup$
    – user76568
    Commented May 10, 2017 at 20:30
  • $\begingroup$ Also $n$ is overloaded and it hurts ;) $\endgroup$
    – user76568
    Commented May 10, 2017 at 20:31
  • $\begingroup$ @Dror I don't understand what '$n$ is overloaded' means. Also, by fixing $\epsilon >0$ and then choosing $N>\frac{4}{\epsilon}$, is my proof then complete? $\endgroup$
    – mrnovice
    Commented May 10, 2017 at 20:32

4 Answers 4

1
$\begingroup$

You should probably specify how you pick $N$ depending on your $\epsilon$, and then use the Archimedean property if you want to be completely rigorous (to show that for all $\epsilon>0$ you find $N$ s.t. $\frac 4 N < \epsilon$). Alternatively, you could just multiply denumerator and nominator by $n^{-3}$ and use some theorems that allow you to compute limits of sums and fractions.

$\endgroup$
0
$\begingroup$

Without getting into the concrete sequence:

You don't really want to show that first two sided implication, as it is just restating the definition.

What does it say, though?

You can read it like:

"If An infinite sequence $a_n$, indexed by $n$ (for poisition), converges to the value $a$ then there is always a position in the sequence for which all of the further elements are closer than whatever distance you wish, to $a$"

So:

  1. The capital $N$ is originally just about the position in the sequence and not the values.

  2. $N$ has to be expressed (If it is by an inequality, then you need more capitals) with $\epsilon$ and with "just numbers".

  3. The mission is to find an experession closer than you originally "intended"

In your concrete example you can just shamelessly divide both the numerator and the denominator by the greatest power of $n$ in it. Then you need to specify many excuses about just usage of arithmetic of limits.

Alternatively you can and should use all the ordinary "re-expression" tricks you have to get a nice and perhaps long ineqality chain that ends and ends obviously. This is essentially what you chose to do. But, you can do it with much less sweat.

$\endgroup$
0
$\begingroup$

If i were you, i would just multiplicate numerator and denominator by $1/n^2$ both, so that: $a_n = (1 + 2/n + 6/n^2)/(n -3/n^2)$. From the defenition of infinite number, we have $2/n$, $6/n^2$ and $-3/n^2$ all equal to 0: $a_n = (1 + 0 + 0)/(n + 0 + 0) = 1/n$ which is 0 from the same defenition.

$\endgroup$
0
0
$\begingroup$

An other proof could use L'Hopital's rule. This should be very quick.

$$ f(n) = \frac{n^2 + 2n + 6}{n^3 - 3} ; f^"(n) = \frac{2}{6n} = \frac{1}{3n}$$

which seems to be much easier to prove that this goes to zero as n goes to infinity...

$\endgroup$
1
  • $\begingroup$ You made $f$ a function of $x$, not $n$. $\endgroup$
    – Harry
    Commented May 10, 2017 at 20:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .