Note: MathJax breaks this page

on most mobile devices.

I am investigating a fix,

but this is low priority.

Suppose we have an annuity-immediate with *n* periods, payment 1 per period, and effective interest rate *i* per period.

The present value of the *m*^{th} payment is equal to \(v^m\), where \(v = \frac{1}{1+i}\): you can think of it is what you would get if you were to pull the payment back through time to time 0, discounting it once per period along the way in a mirror image of what happens with compound interest.

Since an annuity consists of its payments, the present value of an annuity is the sum of the present values of each of its payments. In our case:

\[v^1 + v^2 + … + v^n\]

This is a finite geometric series. Divide through by *v*:

\[v (1 + v + … + v^{n-1}) \]

Apply the formula for the sum of a basic finite geometric series to get

\[v \frac{1 -v^n}{1-v}\]

We can already see the \( 1 – v^n \) we want in the numerator. So what remains to be done is to convert the remaining terms into the *i* in the denominator. Since there is no *i *explicit in the starting formula, what I do at this stage is write out *v* as \(\frac{1}{1+i}\) everywhere except in the term that I want to keep:

\[\color{green}{\frac{i}{1+i}}\frac{1- v^n}{1-\color{green}{\frac{1}{1+i}}}\]

Sidebar: This is not by a long shot the most efficient way to do things, but I am writing a walkthrough, not a proof: proofs are great if you want an efficient demonstration of a proposition, but less so if you are trying to gain your footing. Since proofs get sanitized with the benefit of hindsight, they can be awful for learners when they leave you wondering *“How in the world did you know to do that step?”*

The messiness of the denominator may hinder insight, so we will want to clean it up by combining terms. This requires us first to write out the 1 as \(\frac{1+i}{1+i}\):

\[\frac{1}{1+i}\frac{1- v^n}{\color{green}{\frac{1+i}{1+i}}-\frac{1}{1+i}}\]

And now combine terms:

\[\frac{1}{1+i}\frac{1- v^n}{\color{green}{\frac{1+i-1}{1+i}}}\]

Do some very basic arithmetic, namely \(1 – 1\) (a computation so simple that not even I can screw it up):

\[\frac{1}{1+i}\frac{1- v^n}{\frac{\color{green}{i}}{1+i}}\]

And then multiply straight through to get:

\[\frac{1- v^n}{i}\]

Which is what we were after. Using the notation \(a_{\overline{n}|i}\) to designate the present value of the annuity we started with, we have our basic formula:

\(a_{\overline{n}|i} = \frac{1 – v^n}{i}\)