Please can you help me with this proof

Question

Prove that $\frac{1}{a_\overline{n|} } = i + \frac{1}{s_\overline{n|}}$ all at $i$

This is what I've done so far. Please feel free to change the approach entirely

$$\frac{1}{a_\overline{n|}} = \frac{i}{1-v^n}$$

$$= \frac{i}{1-(1+i)^{-n}} \times \frac{(1+i)^n}{(1+i)^n}$$

$$= \frac{i(1+i)^n}{(1+i)^n-1}$$

$$= \frac{[(1+i) - 1](1+i)^n}{(1+i)^n-1}$$

$$= \frac{(1+i)^{n+1} - (1+i)^n}{(1+i)^n-1}$$

$$= \frac{(1+i)^{n+1}}{(1+i)^n-1} - \frac{(1+i)^n}{(1+i)^n-1}$$

$$= \frac{(1+i)^{n+1} - 1}{(1+i)^n-1} - \frac{(1+i)^n -1}{(1+i)^n-1}$$

$$= \frac{(1+i)^{n+1}-1}{(1+i)^n-1} - 1$$

$$= \frac{(1+i)^{n+1}}{(1+i)^n-1} - \frac{1}{(1+i)^n-1} - 1$$

Now I just feel like I'm running around in circles

Definitions of international actuarial symbols:

$a_\overline{n|}$ is the present value of an annuity of 1 per annum at an effective interest rate of $i$ per annum for $n$ years.

$a_\overline{n|} = \frac{1-v^n}{i}$

$s_\overline{n|}$ is the future value of an annuity of 1 per annum at an effective interest rate of $i$ per annum for $n$ years.

$s_\overline{n|} = \frac{(1+i)^n-1}{i}$

$(1+i)^n$ is when you compound interest for $n$ years

$v^n$ is just discounting backwards

$v^n = (1+i)^{-n}$

Answer 1

Observe that $$\begin{align} \dfrac{1}{a_{\overline{n}|}} = \dfrac{i}{1-v^{n}} &= \left[\dfrac{(1+i)^{n}}{(1+i)^{n}}\right]\left[\dfrac{i}{1-v^n}\right] \\ &= i\left[\dfrac{(1+i)^n}{\left(1+i\right)^{n} - 1}\right] \\ &= i\left[\dfrac{(1+i)^n-1+1}{\left(1+i\right)^{n} - 1}\right] \\ &= i\left[\dfrac{(1+i)^{n}-1}{(1+i)^{n}-1}+\dfrac{1}{(1+i)^{n}-1}\right] \\ &= i\left[1+\dfrac{1}{(1+i)^{n}-1}\right] \\ &= i + \dfrac{i}{(1+i)^{n}-1} \\ &= i + \dfrac{1}{s_{\overline{n}|}}\text{.} \end{align}$$

Answer 2

Starting at this point:

$$\frac{i(1+i)^n}{(1+i)^n-1},$$

try applying the fact that

$$\frac x{x-1} = 1 + \frac 1{x-1}.$$