Interpreting The Weak Law of Large Numbers Given the usual setup, the weak law of large numbers states that for any $\epsilon > 0$
$$
\lim_{n\rightarrow \infty}P(|M_n - \mu| > \epsilon) = 0
$$
According to this author, the interpretation is as follows:

(1) For large values of $n$, (i.e. $n \ge N$ for some $N$) the
  probability that the value of $M_n$ (the sample mean) differs from the
  population mean $\mu$ by more than any given number $\epsilon > 0$
  is 0.
(2) Alternatively, all probability is concentrated in an $\epsilon$
  interval around $\mu$.
(3) Alternatively, almost surely, for large samples, the sample mean is
  within an $\epsilon$ neighborhood of the population mean.

I understand all of this, but am having trouble reconciling this interpretation with the following statement from Exploring Monte Carlo Methods by Dunn and Shultis, which states (notation adapted slightly)

The weak form of the law of large numbers states that, for a specified
  large $N$, the sample mean $M_N$ is likely to be near $\mu$. But
  it leaves open the possibility that cases when $|M_N - \mu|
> \epsilon$, i.e. when the deviation of $M_N$ from $\mu$ is
  outside some small tolerance interval $\epsilon$, can occur an
  arbitrary (even infinite) number of times as $N$ increases; however
  such occurrences happen at infrequent intervals.

Is there a contradiction between these two interpretations?
The first author is saying that for all $n \ge N$, with probability 1, the sample mean is within $\epsilon$ of $\mu$, but the second author says that for $n > N$ the weak law leaves open the possibility we have an infinite number of deviations larger than $\epsilon$. But if this happens an infinite number of times, how can the event occur almost surely?
 A: We say $\{X_n\}_{n=1}^{\infty}$ converges to $0$ "in probability" if for all $\epsilon>0$: 
$$ \lim_{n\rightarrow\infty} Pr[|X_n-0|>\epsilon] = 0 $$
We say $\{X_n\}_{n=1}^{\infty}$ converges to $0$ "with probability 1" if for all $\epsilon>0$: 
$$ \lim_{n\rightarrow\infty} Pr\left[\{|X_n-0|>\epsilon\}\cup\{|X_{n+1}-0|>\epsilon\}\cup\{|X_{n+2}-0|>\epsilon\}\cup\cdots\right] = 0 $$
Of course: 
$$\{|X_n-0|>\epsilon\} \subseteq \{|X_n-0|>\epsilon\}\cup\{|X_{n+1}-0|>\epsilon\}\cup\{|X_{n+2}-0|>\epsilon\}\cup\cdots $$
Thus: 
$$ Pr[|X_n-0|>\epsilon] \leq Pr\left[\{|X_n-0|>\epsilon\}\cup\{|X_{n+1}-0|>\epsilon\}\cup\{|X_{n+2}-0|>\epsilon\}\cup\cdots\right] $$
and so: 
$$ \lim_{n\rightarrow\infty} Pr[|X_n-0|>\epsilon] \leq \lim_{n\rightarrow\infty} Pr\left[\{|X_n-0|>\epsilon\}\cup\{|X_{n+1}-0|>\epsilon\}\cup\{|X_{n+2}-0|>\epsilon\}\cup\cdots\right] $$
You can see that it is easier for the left-hand-side to converge to $0$ than the right-hand-side to converge to $0$. So "convergence with probability 1" implies "convergence in probability," but not vice versa. A standard example to show the reverse does not necessarily hold is this: 
Let $\{X_n\}_{n=1}^{\infty}$ be an infinite sequence of mutually independent random variables such that: 
$$ X_n = \left\{ \begin{array}{ll}
0 &\mbox{ with probability $(n-1)/n$} \\
1  & \mbox{ with probability $1/n$} 
\end{array}
\right.$$
Then $X_n$ converges to $0$ "in probability" because for any $\epsilon>0$ we have $Pr[|X_n-0|>\epsilon] = 1/n$, and so $\lim_{n\rightarrow\infty} Pr[|X_n-0|>\epsilon] = 0$. 
However, it turns out that (with probability 1) we have $X_n=1$ for an infinite number of indices $n$, so $X_n$ certainly does not converge to $0$ with probability 1.  To understand why, fix any arbitrarily large positive integer $M$ and compute the probability that $X_n=0$ for all $n\geq M$: 
$$ Pr\left[\cap_{n\geq M} \{X_n=0\}\right] = \prod_{n=M}^{\infty} \left(\frac{n-1}{n}\right) = 0 $$
The intuition is that, regardless of how large $M$ is, there will (with probability 1) be another index $n$ even larger such that $X_n=1$ (of course, it may take a long time to get there).  Once that occurs, there will again (with probability 1) be an even larger index such that it happens again. And then again, and again, and so on. 
