Prove that the infinite product of (1 + x/n^2) for n = 1 to infinity converges for every x > 0, and find the limit of ln of that product divided by root x as x tends to infinity.

Step-by-step solution

1. 1

   Turn the product into a series. All factors exceed $1$, so the product converges exactly when $\ln P(x)=\sum_{n=1}^{\infty}\ln\!\left(1+\frac{x}{n^2}\right)$ converges. Products are hard to bound directly; the logarithm makes the comparison test available.

2. 2

   Compare with a $p$-series. For $u\ge 0$ the elementary bound $\ln(1+u)\le u$ gives $0\le\ln\!\left(1+\frac{x}{n^2}\right)\le\frac{x}{n^2}.$ Since $\sum 1/n^2=\pi^2/6$ converges, the comparison test forces $\ln P(x)$ to converge for every fixed $x>0$, hence the product converges.

3. 3

   Recognise the closed form. Euler's product for the hyperbolic sine, $\frac{\sinh(\pi z)}{\pi z}=\prod_{n=1}^{\infty}\left(1+\frac{z^2}{n^2}\right),$ applies with $z=\sqrt{x}$, so $P(x)=\frac{\sinh(\pi\sqrt{x})}{\pi\sqrt{x}}.$ At $x=2$ both sides equal $9.56675$, and at $x=9$ both equal $657.397$.

4. 4

   Take logarithms and extract the growth. Using $\sinh w=\tfrac12(e^{w}-e^{-w})$, $\ln P(x)=\pi\sqrt{x}-\ln 2-\ln\!\left(\pi\sqrt{x}\right)+\ln\!\left(1-e^{-2\pi\sqrt{x}}\right).$ The leading term is $\pi\sqrt{x}$; everything else is $O(\ln x)$.

5. 5

   Divide by $\sqrt{x}$ and pass to the limit. $\frac{\ln P(x)}{\sqrt{x}}=\pi-\frac{\ln 2+\ln(\pi\sqrt{x})}{\sqrt{x}}+o(1)\longrightarrow\pi.$ Equivalently, the sum is a Riemann sum with mesh $1/\sqrt{x}$ for $\int_0^{\infty}\ln(1+t^{-2})\,dt=\pi$. Direct summation at $x=10^{6}$ gives $3.1278$, converging slowly toward $\pi$ as the $\ln x/\sqrt{x}$ correction predicts.