This article shows that
and thus that the function
is bounded above and below.
If is an integer, write where the are distinct primes. Then
Proof. Note that the number of multiples of is simply . But that does not result in since some of those numbers may have additional factors of . So divide each by ; then clearly
The result follows inductively. Note that the sum is actually finite.
If , all distinct, then .
Proof. Consider . Choose such that ; it suffices to show .
Now, in general, if , then or . So the sum above has at most terms (since ), each of which is either or , and thus .
Proof. Choose such that , and write . Each by the preceding lemma, and there are at most terms on the right-hand side. Thus
Summing from to , we get
But since , so
so and thus
Proof. Note that if is prime, , then divides since occurs in the numerator but not in the denominator. Thus divides as well, and thus
Taking logs, we get
Let . Then
Note that is monotonically increasing, so if we choose with , then
|Date of creation||2013-03-22 16:23:51|
|Last modified on||2013-03-22 16:23:51|
|Last modified by||rm50 (10146)|