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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

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

Clarinetist
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StephanCasey
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  • If you'd define all your signs and symbols that'd help a lot. – Timbuc Oct 27 '14 at 13:45
  • I tried. I hope you can help me if you don't even know international actuarial symbols – StephanCasey Oct 27 '14 at 13:58
  • Funny thing: I was a lecturer of mathematics in the faculty of actuarial science for more than 3 years, yet I never learned (honestly, I didn't get interested, either) all that financial lingo. Sorry. BTW, for anyone knowing this, you did not define $;V^n;$ . – Timbuc Oct 27 '14 at 14:02
  • Oh sorry I'll do it now – StephanCasey Oct 27 '14 at 14:06
  • @StephanCasey - I have edited your post. Note that there is a difference between $V^n$ and $v^n$. I will respond as soon as I can. – Clarinetist Oct 27 '14 at 14:12

2 Answers2

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

Clarinetist
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Starting at this point:

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

try applying the fact that

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

David K
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