## Expression of Pi as Infinite Product

Here's a very interesting problem posted by Paul. This is an advanced problem, so don't feel bad if you don't understand the fi rst time you read this.You should had mastered integration by parts. Our goal is to prove the following formula:

With this we can express pi as a product (more preciselly, the limit of a product). And here again we'll see an interesting application of integration by parts. First of all, we'll prove the following integrals using integration by parts:

To prove this we'll use a similar technique we used to prove this recurrence formula. We'll defi ne the function:

For n=0 we have:

And then, if n>0, we have:

We'll integrate this by parts, making:

So, by integration by parts:

And there we have our recurrence formula:

We can start calculating explicitly the rst integrals to see what is the pattern. We already shown what is J(0):

And there the pattern is clear. The general formula would be:

And J(n) is our integral!

To prove the second integral we proceed in a similar way. First we de fine another function:

And calculate the fi rst value:

And then we use integration by parts to nd a recurrence formula:

And our recurrence formula is:

Solving for K(n):

And let's calculate the fi rst values:

And the pattern is clear. The general formula is:

Now we have our formulas. We can proceed to the proof of the infi nite product expression of pi. This is the easy part. Let's note that:

And if we solve for pi/2:

Now, if we take the limit as n approaches in nity to both sides of the equation, we should get our result. Should we? What is left for us to prove is that:

Or, what is the same:

To do this, we'll use the squeeze theorem. So, rst of all, we need to form some inequalities. We'll establish the following inequalities:

From this, using the squeeze theorem, and noting that the limit in the first term of the inequality is one, we get the result we want. First, let's prove that:

Let's express these as integrals:

And noting that:

We now use the property of the de nite integral that says that the integral of one function that is less than other is itself smaller:

So, we have:

Similarly, let's prove that:

Taking inverses:

And multiplying both sides by J(n):

But, from our recurrence formula:

So, we get the double inequality:

It is easy to calculate the limit as n approaches in nity of left member:

So, by the squeeze theorem, we have that:

And taking the inverse:

From the formula we arrived earlier, and taking limits on both sides:

And we proved a non-trivial result(!!). This is what is known as Wallis Product. It is a pretty amazing formula, and we proved using just a few things we learned in calculus: integration by parts, some properties of de nite integrals and the squeeze theorem.