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