Where does this characterization of an annuity immediate come from? 
I'm looking through my notes, and I don't see anywhere that an annuity immediate can be defined as $a_n = \frac{1}{a(1)} + \frac{1}{a(2)} + \cdots + \frac{1}{a(n)}$.
I've always seen it as $a_n = v + v^2 + \cdots + v^n$.
Any references for why this author's characterization is true?
 A: Note that $\delta$ is constant if and only if $i$ is constant, and if $i$ is constant, we define $v = \dfrac{1}{1+i}$.
So actually, the formula $$a_{\overline{n}|} = \sum\limits_{j=1}^{n}v^j\tag{1}$$ is only true for constant $i$, and therefore constant $\delta$.
One problem I've had with actuarial texts is that they seem to be ambiguous with their notation. Personally, I don't find $a_{\overline{n}|}$ to be appropriate notation for this problem since this usually implies that $i$ is constant. But the SOA has done that in their problem, so I guess there's no use arguing against that.
Let's try to extend $a_{\overline{n}|}$ to a non-constant force of interest. We will do this by starting with how the idea of present value was originally motivated, but using variable force of interest instead. 
Suppose you have a cash flow of $P$ invested immediately growing at force of interest $\delta_t$, $t > 0$. You should know that at any time $t > 0$, $P$ grows to $$A(t) = Pa(t) = P\exp\left(\int\limits_{0}^{t}\delta_s \text{ d}s\right)\text{,}$$
which is the future value of $P$ at time $t$. To solve for the present value (which is obviously $P$), we have
$$P = \dfrac{A(t)}{\exp\left(\int\limits_{0}^{t}\delta_s \text{ d}s\right)} = \dfrac{A(t)}{a(t)}\text{.}$$
Now the above is the present value of one cash flow at time $t$. 
$a_{\overline{n}|}$ assumes we have cash flows (or future values) of $1$ at integer times $1, 2, \dots, n$. So at time $1$, we have a present value of $\dfrac{1}{a(1)}$; at time $2$, we have $\dfrac{1}{a(2)}$, and so forth. Thus,
$$a_{\overline{n}|} = \sum\limits_{j=1}^{n}\dfrac{1}{a(j)}\tag{2}$$
as a more general definition. In the case of constant $i$, we have $a(j) = (1+i)^{j}$, so that $$\dfrac{1}{a(j)} = \dfrac{1}{(1+i)^{j}} = v^{j}$$ by definition, so $(2)$ agrees with $(1)$ .
