# Ask Uncle Colin: An Infinite Product

Dear Uncle Colin,

I have to show that $\Pi_1^\infty \frac{(2n+1)^2 - 1}{(2n+1)^2} > \frac{3}{4}$. How would you do that?

Partial Results Obtained Don’t Undeniably Create Truth

Hi, PRODUCT, and thanks for your message!

That’s a messy one. I can see two reasonable approaches: one is to take the whole thing into log-world; the other is to expand and see if the factorials tell us anything.

### Log-world

If we take the logarithm of the infinite product (let’s call it $P$, it turns into an infinite sum:

$\ln(P) = \Sigma_1^\infty \ln\left( \frac{(2n+1)^2 - 1}{(2n+1)^2}\right)$

We can rewrite each term as $\ln \left( 1 - \frac{1}{(2n+1)^2} \right)$.

Now, what can we say about a lower bound for each of those terms?

#### An aside into $e$-world

We know that $e^x \ge 1+x$ for all $x$, so $x \ge \ln(1+x)$.

A consequence of that is that $x-1 \ge \ln(x)$ (simply replacing every $x$ with $x-1$).

Similarly, replacing $x$ with $\frac{1}{x}$ give $\frac{1}{x}-1 \ge \ln\left(\frac{1}{x}\right)$ - so, being careful with the inequality direction, $\ln(x) \ge 1 - \frac{1}{x}$.

But we can go further: we’re particularly interested in $\ln\left(1 - \frac{1}{x}\right)$.

Running through the algebra, this ends up giving $\ln\left(1 - \frac{1}{x}\right) \ge -\frac{1}{x-1}$.

#### Back to the logs

This lower bound is good: it tells us that each term - I’m going to call it $T_n$ - has the property $T_n = \ln \left( 1 - \frac{1}{(2n+1)^2} \right) \ge -\frac{1}{(2n+1)^2-1}$.

That bottom sends my spidey-senses a-tingling: it’s a difference of two squares!

So, $T_n \ge -\frac{1}{4n(n+1)}$

Tingle on, spidey-senses: that’s a partial fraction!

$T_n \ge \frac{1}{4n+4} -\frac{1}{4n}$.

And now I spy a telescoping sum!

$\sum_{n=1}^{\infty} T_n \ge \left(-\frac{1}{4} + \frac{1}{8}\right) + \left(-\frac{1}{8} + \frac{1}{12}\right) + \left( -\frac{1}{12} + \frac{1}{16}\right) + \dots$. Everything except the initial $ -\frac{1}{4}$ cancels out!

So! $\ln(P) = \sum_{n=1}^{\infty} \ge -\frac{1}{4}$, and $P \ge e^{-\frac{1}{4}}$. Since $e^x > 1 + x$, we have $P \ge e^{-\frac{1}{4}} \gt \frac{3}{4}$ as required!

### What, you want more?!

An alternative method is to consider the product directly:

$P = \br{1 - \frac{1}{9}} \br{1 - \frac{1}{25}} \br{1 - \frac{1}{49}} \dots$

$P > 1 - \sum_{n=1}^{\infty} \frac{1}{(2n+1)^2}$ (I haven’t justified this, but it’s true).

But what’s the sum of the reciprocals of the odd squares?

#### Bustle into Basel

Now, we know that the sum of the reciprocals of *all* of the squares is $\frac{\pi^2}{6}$, because that’s the Basel problem.

We can deduce from that that the sum of the reciprocals of all of the *even* squares is $\frac{\pi^2}{24}$, because each term is a quarter of the above.

So, the sum of the reciprocals of the odd squares is $\frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{\pi^2}{8}$.

However, that’s for all the odd squares - we’re starting at 3 rather than 1, so we know that $P > 1 - \br{\frac{\pi^2}{8} - 1} = 2 - \frac{\pi^2}{8}$.

We know that $\pi^2 < 10$, so $\frac{\pi^2}{8} < \frac{5}{4}$, which means that $2 - \frac{\pi^2}{8} > \frac{3}{4}$ - and we’re done!

Phew.

Hope that helps!

- Uncle Colin