Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}\sqrt{\frac{n^2+2}{n^2+1}}$, and evaluate it.
Explanation of Method
Since $\frac{n^2+2}{n^2+1}=1+\frac1{n^2+1}\sim1+\frac1{n^2}$, comparison with $\sum1/n^2$ shows convergence. To find the exact value, this product is a ratio of two instances of the classical Weierstrass sine-hyperbolic product $\sinh(\pi x)=\pi x\prod_{n=1}^\infty\bigl(1+x^2/n^2\bigr)$, evaluated at $x=\sqrt2$ and $x=1$.
- Show convergence via comparison with $\sum1/n^2$.
- Recognize $\prod(n^2+x^2)/n^2=\sinh(\pi x)/(\pi x)$ and apply it twice.
- Divide the two instances and take the square root.
Step 1 · Convergence
$$ \frac{n^2+2}{n^2+1}=1+\frac1{n^2+1}, $$ and since $\frac1{n^2+1}\sim\frac1{n^2}$, comparison with the convergent $p$-series $\sum1/n^2$ shows $\sum\ln\sqrt{\frac{n^2+2}{n^2+1}}$ converges, so the product converges to a positive finite limit.
Step 2 · The Weierstrass hyperbolic-sine product
The classical identity $$ \sinh(\pi x)=\pi x\prod_{n=1}^\infty\left(1+\frac{x^2}{n^2}\right) \quad\Longleftrightarrow\quad \prod_{n=1}^\infty\frac{n^2+x^2}{n^2}=\frac{\sinh(\pi x)}{\pi x} $$ applies with $x=\sqrt2$ and $x=1$: $$ \prod_{n=1}^\infty\frac{n^2+2}{n^2}=\frac{\sinh(\pi\sqrt2)}{\pi\sqrt2}, \qquad \prod_{n=1}^\infty\frac{n^2+1}{n^2}=\frac{\sinh\pi}\pi. $$
Step 3 · Divide and take the square root
$$ \prod_{n=1}^\infty\frac{n^2+2}{n^2+1}=\frac{\sinh(\pi\sqrt2)/(\pi\sqrt2)}{\sinh(\pi)/\pi}=\frac{\sinh(\pi\sqrt2)}{\sqrt2\,\sinh\pi}. $$ $$ \boxed{\prod_{n=1}^{\infty}\sqrt{\frac{n^2+2}{n^2+1}}=\sqrt{\frac{\sinh(\pi\sqrt2)}{\sqrt2\,\sinh\pi}}\approx1.613206} $$ (Numerical check: direct product to $10^6$ terms gives $\approx1.613206$ — matching the closed form to 6 significant figures; the convergence is slow, $O(1/N)$, since the tail behaves like $\sum1/n^2$.)
Summary Table
| Property | Value |
|---|---|
| Closed form | $\sqrt{\dfrac{\sinh(\pi\sqrt2)}{\sqrt2\,\sinh\pi}}$ |
| Numerical value | $\approx1.613206$ |
Additional Notes
- Correction: a draft attempt guessed the limit was $\sqrt e\approx1.648721$ based on a small handful of loosely estimated partial products. Careful computation shows the partial products actually stabilize near $1.6132$, not $1.6487$ — a genuine numerical error, not a slow-convergence artifact (verified to $10^6$ terms). The exact closed form, obtained via the Weierstrass product for $\sinh$, is $\sqrt{\sinh(\pi\sqrt2)/(\sqrt2\sinh\pi)}\approx1.613206$, matching the true numerics precisely.
- This is the same Weierstrass product mechanism that underlies the classical evaluation $\prod_{n=1}^\infty(1+1/n^2)=\sinh(\pi)/\pi\approx3.676$ — indeed that is exactly the $x=1$ building block used in Step 2 above.
- Unlike Problems 360–363 and 365–366, which telescope to elementary rational or exponential closed forms, this product only resolves in closed form because it is expressible as a ratio of two Weierstrass products for $\sinh$ — a genuinely different (though still classical) mechanism.