Weak law of large numbers - redundant? I might be missing something basic - but it appears that the strong law of large numbers covers the weak law. If that case, why is the weak law needed?
 A: The mathematical formulations of the "Strong" and "Weak" Laws of Large Numbers look somewhat similar. Yet, the two Laws are quite different in nature :
The Weak Law never considers infinite sequences of realizations of a random variable. It only states that imbalanced sequences are less likely to occur as one considers longer sequences.
On the other hand, the Strong Law considers only infinite sequences of realizations of a random variable, and more precisely, the set of these infinite sequences. It states that the set of imbalanced sequences has probability 0 in a sense that generalizes the concept of "set of measure 0".
It can be shown that the Strong Law implies the Weak Law, which can therefore be regarded as a consequence of the Strong Law.
The converse is, however, wrong : it is possible to exhibit sequences of r.v.s following the Weak Law, but not the Strong one. So the terms "Weak" and "Strong" are indeed justified. For example, Let your sequence be i.i.d. with density
$f_X(x)=x^{-2}I(x>1)$
You can obtain a WLLN but not a SLLN, due to the Borel-Cantelli lemma. 
A: The most general case of the Weak Law of Large Numbers does not even require the existence of first moments. Therefore, it holds under conditions/assumptions more general than the conditions/assumptions required for the Strong Law of Large Numbers (existence of first moments).
Allow me to quote for you the relevant results from Durrett, Probability: Theory and Examples (4th edition), so you can see the truth of the above statement for yourself.

(p.60) Theorem 2.2.7 Weak Law of Large Numbers Let $X_1, X_2, \dots$ be i.i.d. with $$x \mathbb{P}(|X_i|>x) \to 0 \quad as \quad x \to \infty \quad (for\,all\,i=1,2,\dots) $$ Let $S_n = X_1 + \dots + X_n$ and $\mu_n = \mathbb{E}[X_1 1_{|X_1 \le n|}]$ [edit was to the left - remove stuff in [ ] ]. Then $S_n/n - \mu_n \to 0$ in probability.

The condition, for each $X_i$ in the sequence of random variables, that $x \mathbb{P}(|X_i| > x) \to 0$ as $x \to \infty$ is strictly weaker than the existence of first moments -- i.e., there exist i.i.d. sequences of random variables which satisfy this condition but which do not have finite first moments. For an example, see the previous answer above.

(p.73) Theorem 2.4.1. Strong Law of Large Numbers Let $X_1, X_2, \dots$ be pairwise independent identically distributed random variables with $\mathbb{E}|X_i| < \infty$ (for all $i = 1, 2, \dots$). Let $\mathbb{E}X_i = \mu$ and $S_n = X_1 + \dots + X_n$. Then $S_n/n \to \mu$ almost surely as $n \to \infty$.

Theorem 2.4.5. on p.75 is the Strong Law for the case that the first moment exists but is not finite.
Both results (the Weak Law of Large Numbers and the Strong Law of Large Numbers) are a lot easier to prove if/when we assume that the random variables have finite variance (second moments), but such an assumption is unnecessary for both results.
So, in conclusion, the Weak Law of Large Numbers is not redundant, because although its conclusion is weaker than that of the Strong Law of Large Numbers, it is true "more often" (i.e. under more general conditions) than the Strong Law of Large Numbers. So even when the Strong Law doesn't hold, the Weak Law may still hold.
