Sum of prime factors

Question

If $x$ is an odd prime, is it true that the sum of the prime factors of $x + 1$ is less than $x$?

If so, then this would give a nice way of constructing a "jumpy" sequence that converges to $0$, namely, let the $n$th term be $1$ divided by the sum of the prime factors of $n + 1$. Then if $n + 1$ is an odd prime, then the $n$th term is less than the $n + 1$th term, so that we cannot say, "We're always getting closer to $0$, every step of the way.", but the sequence does in fact converge to $0$.

This illustrates why the rigorous definition of convergence is needed: convergence of bounded monotonic sequences can be handled off-handedly, but not so the general case.

edit (4.Sep.2013, CST, MERCA):

In fact, non-monotonic convergence to zero is a familiar physical fact, and there are some expressions that capture this notion. Here are three:

"flash-in-the-pan"

"dead-cat bounce"

"death-rattle"

Can anyone come up with any others?

Calculus teachers could perhaps make use of these expressions in motivating the formal definition of limit.

Answer 1

Lemma: The sum (counting multiplicity) of the prime factors of any positive integer $n$ is $\le n$.

The lemma is proved at the end of this post. Using the lemma, we find that the sum of the prime factors of $2m$ is $\le 2+m$.

Let $x$ be an odd number, and let $2m=x+1$. The sum of the prime factors of $x+1$ is $\le 2+\frac{x+1}{2}$. And we have $$2+\frac{x+1}{2}\lt x$$ if $x\gt 5$.

Proof of Lemma: We use (strong) induction. Let $S(n)$ be the sum of the prime factors of $n$. Suppose $S(k)\le k$ for all positive $k\lt n$. We want to show that $S(n)\le n$. The result is obvious if $n$ is prime. If $n$ is not prime, let $n=ab$, where $a,b\lt n$. Then $S(n)=S(a)+S(b)$. By the induction assumption $S(a)+S(b)\le a+b$. But $a+b\le ab$. This completes the proof.