Meaning of NOT all but finitely often Can someone clarify for me the meaning of the statement "NOT all but finitely often"?
It's driving me crazy. I'm not able to break it up.
Thanks.
 A: A property holds for "all but finitely many $x$" is the set of $x$ on which it fails is finite. 
The negation of the above is: the set  on which the property fails is infinite. In other words, it fails infinitely often. 
A: The phrase not all but finitely often means for some set $A$ and property $P(x)$, it is not the case that $P(x)$ holds for all $x \in A$, but for some finite set $B \subseteq A$, it is the case that $P(x)$ holds for all $x \in B$. In order to demonstrate that $P(x)$ holds for not all $x \in A$ but only many $x \in A$, you must show that $P(x)$ is false for some $x \in A$ but true for some finite subset of $A$ (which may be empty).
A trivial example of a property $P(n)$ and set $A$ for which $P(n)$ holds of only finitely many $x \in A$ is $P(n) \equiv n < 5$ and $A = \mathbb{N}$. Clearly $P(n)$ fails for all $n \geq 5$, so $P(n)$ fails to hold for all $n \in \mathbb{N}$. However, $P(n)$ holds for the finite set $B = \{ n \in \mathbb{N} : n < 5\}$, so $P(n)$ is true for not all but only finitely many natural numbers.
Another example is $P(n) \equiv \int x^n \neq x^{n + 1}/(n + 1) + C$ and $A = \mathbb{N}$. Clearly, $P(n)$ fails whenever $n \neq -1$, so $P(n)$ is not true for all natural numbers $n$. However, $P(n)$ is true whenever $n = -1$, since $\int x^{-1} = \log|x| + C$. Thus, the set of natural numbers for which $P(n)$ is true is finite. So $P(n)$ holds of not all but only finitely many natural numbers.
