In symbols, my question is asking the following. Let:
$a_{\overline{n|}} = {}$ present value of an immediate annuity payable for n units of time
$\ddot{a}_{\overline{n|}}^{(p)}={}$present value of an annuity due payable p times per year for n units of time
Then, is the following statement correct? $$\ddot{a}_{\overline{n|}}^{(p)}=1+a_{\overline{n-1|}}^{(p)}$$
(Apologies for the bad notation. I don't know how to enter in actuarial notation)
Note that $\ddot{a}_{\overline{n}|}^{(p)}$ describes an annuity-due which pays $\dfrac{1}{p}$ for $p$ times per year for $n$ years. I mention this, because most people forget that this symbol does not represent payments of $1$, but $\dfrac{1}{p}$. A similar principle applies to $a_{\overline{n}|}^{(p)}$.
From this understanding, we obtain - assuming a $p $thly interest rate of $j$ with $v_j = \dfrac{1}{1+j}$ with $j$ satisfying $(1+j)^p = 1+i$ with $i$ being the annual interest rate, $$\ddot{a}_{\overline{n}|}^{(p)} = \dfrac{1}{p}\sum_{k=0}^{np-1}v_j^k$$ and $$a_{\overline{n-1}|}^{(p)}=\dfrac{1}{p}\sum_{k=1}^{p(n-1)}v_j^k = \dfrac{1}{p}\sum_{k=1}^{np-p}v_j^k\text{.}$$
Thus, the following is clear:
1) For $p = 1$, these two sums are equal.
2) For $p > 1$, notice that $$\begin{align} a_{\overline{n-1}|}^{(p)} &= \dfrac{1}{p}\sum_{k=1}^{np-p}v_j^k \\ &= \dfrac{1}{p}\sum_{k=1}^{np-1}v_j^k +\dfrac{1}{p}(v_j^{np-2}+v_j^{np-3}+\cdots+v_j^{np-p}) \\ &= \ddot{a}_{\overline{n}|}^{(p)}-\dfrac{1}{p}+\dfrac{1}{p}v_j^{np-2}(1+v_j^{-1}+\cdots+v_j^{2-p}) \\ &= \ddot{a}_{\overline{n}|}^{(p)}-\dfrac{1}{p}+\dfrac{1}{p}v_j^{np-2}\left[1+(1+j)+\cdots+(1+j)^{p-2}\right] \\ &= \ddot{a}_{\overline{n}|}^{(p)}-\dfrac{1}{p}+\dfrac{1}{p}v_j^{np-2}s_{\overline{p-1}|j}\text{.} \end{align}$$ It's been about six years since I've seen this material, so this is my best attempt.
