No, it is not infintesimal. It is irrational.
Proof:
Choose an integer N M. It follows then that . Define a polynomial . Expanding the polynomial one finds that , where each is an integer. For integers k such that , one also finds that . Now notice that since and . Now suppose this is true for an integer k, i.e. . Now
.
So the formula is true for all nonnegative integers by the principle of mathematical induction. Now for values of k such that each term in the expansion of and . Now for values of k such that the only term which does not contain a positive integer power of x is the first. This is the case when n=k. This means that , an integer since . It follows then that , also an integer.
Now define another function
.
It now follows that and are both integers and thus that is an integer. Now define . It follows then that
.
.
Now
, and so
. Also
.
Notice that since f(x) is a polynomial of degree 2N, for all x. Thus
. So
.
Now since g(x) is continuous on , and exists on , by the Mean Value Theorem, there exists a c such that . Now , and
. So , and thus
.
.
Since and also 0 < 1-c < 1. Since , it follows that .
It now follows that and so . Since is an integer, there must exist an integer between 0 and 1.
Thereby, reaching a contradiction, the theorem is proved. Q.E.D.