Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've just started Real Analysis. In the textbook (Real Analysis and Applications, by Davidson and Donsig) they have defined the limit of a sequence. I am working on one of the provided exercises. No suggested solutions are provided for any questions, so I am looking for help checking my work to make sure I understand what I am doing. The exercise is as follows.

Compute the limit. Then, using $\epsilon =10^{-6}$, find an integer $N$ that satisfies the limit definition. $$\lim_{n \to \infty} \frac{1}{ln(ln(n))}$$

Firstly, I just want to address the limit definition. In the text, they define a real number $L$ to be the limit of a sequence of real numbers $(a_n)^{\infty}_{n=1}$ if for every $\epsilon > 0$ there is an integer $N=N(\epsilon)>0$ such that

$$|a_n - L|< \epsilon$$

for all $n\geq N$. Now for my informal interpretation.

I believe the idea is that after some point the distance between $a_n$ and the limit $L$ can be made arbitrarily small by choosing a sufficiently large $N$. But why does $N$ have to be a function of $\epsilon$? Any clarification of the definition would be appreciated.

Now, for the exercise. I know the divisor of the fraction of the given sequence approaches infinity as $n$ approaches infinity and as such the sequence approaches zero as $n$ approaches infinity. So I claim that $L=0$. I observe that

$$\left | \frac{1}{ln(ln(n))} - 0 \right | = \frac{1}{ln(ln(n))}$$

So now I need an $N$ such that, for all $n \geq N$, I get $|a_n - 0|<10^{-6}$. Now, I am not sure if my next steps are correct. I need an $N$ such that

$$\frac{1}{ln(ln(N))}<10^{-6}$$

So I just solve for N in the equality, which yields

$$N > e^{e^{1000000}}$$

Hence, for any $n \geq N$,

$$\frac{1}{ln(ln(n))}<10^{-6}$$

Do I find the closest integer to $e^{e^{1000000}}$? Any clarification would be appreciated.

share|improve this question
    
It looks to me like you are done. You could compute $N$, but keep in mind it doesn't hurt for $N$ to be higher. It doesn't have to be the minimal one over $e^{e^{1000000}}$ –  rschwieb Jan 10 '13 at 21:46
    
You can just write your integer as $\lceil e^{e^{1000000}} \rceil$ to make it clear you're talking about an integer. Other than that you have found a value that works and so you are done. –  AvatarOfChronos Jan 10 '13 at 21:49
    
I suspect that computing $e^{e^{1000000}}$ would be a major project, requiring quite a bit of programming skill. It doesn't make sense to try to compute the exact value in this context at all. –  Henry B. Jan 10 '13 at 22:33
1  
it's a very big number, even computers are afraid to compute it. –  Santosh Linkha Jan 10 '13 at 22:43

1 Answer 1

up vote 0 down vote accepted

I think your confusion arises from the phrase "there exists an $N = N(\epsilon)$. You already showed why $N$ has to be a function of $\epsilon$: intuitively, the smaller the epsilon, the larger the $N$ has to be in order for the inequality to be satisfied.

You could just say then: choose $N > e^{e^{\epsilon^{-1}}}$ for $N(\epsilon)$; if you show such an $N$ exists then this implies that such a function exists. You don't have to exhibit a specific one, unless you want to (or asked on a question).

But, if you want to be explicit, you could use:

  • $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil$ where $\lceil - \rceil$ denotes the least integer greater than its argument, as you thought
  • $N(\epsilon) = \lceil e^{e^{\epsilon^{-1}}}\rceil + 8434$
  • $N(\epsilon) = \lceil 9434\pi e^{e^{\epsilon^{-1}}}\rceil$

See, all of them work, as long as the function gives you an integer sufficiently larger so that the $|a_n - L| < \epsilon$. However, I should stress once more that as long as you show that there exists such an integer for every $\epsilon$, this already shows that the function exists. (Modulo your philosophy on mathematics; you might actually need to construct the function for your peace of mind.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.