Bounds on tail of distribution by sum of tails 
From what I can see, the tail of the convolution is greater than the stuff approximated by the sum of tails. should this not imply the liminf is greater or equal to one?
 A: This picture shows the idea.

It depicts the situation with two random variables, plotting $Z_1$ on the horizontal axis and $Z_2$ on the vertical axis.  Region IV is the set where $Z_1 \gt z$ and $Z_2 \le z$.  Provided we pick $z$ large enough, we can guarantee most of the probability of $Z_1$ is in the interval $[0,z]$ and most of the probability of $Z_2$ is in this interval, too.  These are the quantities $F_1(z)$ and $F_2(z)$.  Both lie within an arbitrary narrow range $[1-\epsilon]$. The complementary probabilities are $\bar F_i(z) = 1 - F_i(z)$.
Because $Z_1$ and $Z_2$ are independent, the probability of any rectangle in this picture is the product of the marginal probabilities.  Specifically, the chance that $(Z_1,Z_2)$ is in region $IV$ is
$$\Pr(IV) = \bar {F}_1(z) F_2(z) \ge \bar {F}_1(z) (1-\epsilon ).$$
The same argument, with $Z_1$ and $Z_2$ switched, establishes that the chance of region $II$ is
$$\Pr(II) = \bar{F}_2(z) F_1(z)\ge \bar {F}_2(z) (1-\epsilon ).$$
Now the event $Z_1 + Z_2 \gt z$ consists of regions $I, II, III,$ and $IV$.  The chance that $Z_1+Z_2$ exceeds $z$ obviously equals the sum of the chances of each of the regions (provided we are careful to define their boundaries so they have no overlap).  Consequently
$$\eqalign{
\overline{(F_1 \star F_2)}(z) &= \Pr(Z_1 + Z_2 \gt z) \\
&= \Pr(I)+\Pr(II)+\Pr(III)+\Pr(IV) \\
&\ge \Pr(II) + \Pr(IV) \\
&\ge (1-\epsilon )(\bar{F}_1(z) + \bar{F}_2(z)).
}$$
If $\bar{F}_1(z) + \bar{F}_2(z)=0$ there's nothing to show.  Otherwise, dividing yields
$$\frac{\overline{(F_1 \star F_2)}(z)}{\bar{F}_1(z) + \bar{F}_2(z)} \ge 1-\epsilon $$
for any $\epsilon \gt 0$ provided only that $z$ is chosen sufficiently large (depending on how small $\epsilon $ is).
By definition, this guarantees that the limit infimum of the fraction can be no less than $1-\epsilon $ for all $\epsilon \gt 0$.  But that is equivalent to saying the limit infimum is at least $1$.
The same idea applies directly to more than two random variables (but not necessarily to an infinite number).  The general result also follows, almost immediately, by induction.
