Is $\prod_{p\ge 3} \left(1+p\left(\frac{1}{(p-1)^2}-\frac{1}{p^2}\right)\right) \ge 2$ where the product runs over all odd primes?

Question

We know that $\frac{1}{p_{k-1}^2}-\frac{1}{p_k^2}\ge \frac{1}{(p_{k}-1)^2}-\frac{1}{p_k^2}$ but I do not see further...

Answer 1

Though it's not obvious, the product is actally less than $2$. A typical way to deal with products like this is to turn them into sums by take logs. I.e. if $P$ is the product over primes, then repeatedly using the property $\log(ab) = \log(a) + \log(b)$ gives

$$\log(P) = \sum_{p\ge 3}{\log\left(1+p\left(\frac{1}{(p-1)^2}-\frac{1}{p^2}\right)\right)} = \sum_{p\ge 3}{\log\left(1+\frac{2p-1}{p(p-1)^2}\right)}$$

From here, primes are too difficult to deal with, so we can instead sum over all odd numbers starting at $3$.

$$\log(P) \lt \sum_{n=1}^{\infty}{\log\left(1+\frac{2(2n+1)-1}{(2n+1)(2n+1-1)^2}\right)}$$

$$= \sum_{n=1}^{\infty}{\log\left(1+\frac{4n+1}{4n^2(2n+1)}\right)}$$

At this point we can use the bound $\log(1+x) \lt x$ to simplify the sum, however, we'll only do this for $n\ge 2$ to minimise errors. The upper bound now looks like:

$$\log\left(1+\frac{5}{12}\right)\ +\ \sum_{n = 2}^{\infty}{\frac{4n+1}{4n^2(2n+1)}}\ \lt\ \log\left(1+\frac{5}{12}\right)\ +\ \sum_{n = 2}^{\infty}{\frac{1}{2n^2}}$$

$$= \log\left(\frac{17}{12}\right) + \frac{\pi^2}{12} - \frac{1}{2}\ =\ 0.67077\dots\ \lt\ \log(2)$$

Thus, $P\lt 2$ is obtained by exponentiating both sides.