No, it is not infintesimal. It is irrational.
Proof:
Suppose, by way of contradiction, that 
is rational. Then 
must also be rational. Thus 
where a,b 
Z, b 
0. Since 
, it follows that there exists an M > 0 such that if n 
M, 
, or 
.
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.