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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.

Delightful_Richard
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  • 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
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    it's a very big number, even computers are afraid to compute it. – S L Jan 10 '13 at 22:43

1 Answers1

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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.)