Thread: Limit of an interesting sequence

Problem: Fix and let m,n_0)=1\right\}" alt="E_n=\left\{m\leqslant nm,n_0)=1\right\}" /> (i.e. the set of all where and are coprime). Then, define . Compute .

Remark: I recently solved this problem on my blog, if you have seen it I would ask you refrain from answering.

What is card?

Originally Posted by Also sprach Zarathustra

What is card?

Cardinality

Originally Posted by Drexel28

Problem: Fix and let m,n_0)\right\}" alt="E_n=\left\{m\leqslant nm,n_0)\right\}" /> (i.e. the set of all where and are coprime). Then, define . Compute .

Remark: I recently solved this problem on my blog, if you have seen it I would ask you refrain from answering.

I guess you mean m,n_0)=1\right\}" alt="E_n=\left\{m\leqslant nm,n_0)=1\right\}" />.

Isn't it just the product of the where ranges over the prime divisors of , i.e. the proportion of integers relatively prime to ?

Originally Posted by Bruno J.

I guess you mean m,n_0)=1\right\}" alt="E_n=\left\{m\leqslant nm,n_0)=1\right\}" />.

Isn't it just the product of the where ranges over the prime divisors of , i.e. the proportion of integers relatively prime to ?

Yes, that is what I mean. And I don't know, is it? >:]

Welcome back on MHF by the way

Originally Posted by Drexel28

Yes, that is what I mean. And I don't know, is it? >:]

Yes, it is. This is barely a challenge problem! (*with instead of )

Originally Posted by Bruno J.

Yes, it is. This is barely a challenge problem! (*with instead of )

Ah, ****. My bad then. You see, the one subject I am truly ignorant of is number theory. That said, the way I devised of doing it was probably much more difficult than the number theoretic approach. What I did was (sketch):

Spoiler:

If then . Thus, if _j\mid m\right\}" alt="A_{j,n}=\left\{m\leqslant n_j\mid m\right\}" /> we see that

where the square cup just means that forms a partition of . Thus, . But, by the inclusion exclusion principle

where . Thus, noticing that any subcollection of is a family of coprime numbers we see that

_s\mid m\text{ for all }s\in S\right\}=\left\{m\leqslant n:\prod_{s\in S}p_s\mid m\right\}" alt="\displaystyle \bigcap_{s\in S}A_{s,n}=\left\{m\leqslant n_s\mid m\text{ for all }s\in S\right\}=\left\{m\leqslant n:\prod_{s\in S}p_s\mid m\right\}" />

From where simple reasoning leads us to

Thus,

From where it quickly follows that

At least we know now that

Originally Posted by Bruno J.

Yes, it is. This is barely a challenge problem! (*with instead of )

OH MY GOD! Is the number theoretic way to do it this:

Spoiler:

Let

Notice that . Thus, one can easily see that . It easily follows then by induction that . But, evidently by definition . Thus, . So, notice that is increasing since . Then, if denotes reduction that

But, notice that since that are both integers. Thus, our previous discussion applies and tells us that

and

Notice then though that clearly

So, that by the squeeze theorem it follows that

My mind is blown.