Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}\frac1{n^\alpha\sin\frac1{n^\alpha}}$.
Explanation of Method
Writing $y=1/n^\alpha\to0$, the Taylor expansion $\sin y=y(1-y^2/6+\cdots)$ turns each factor into $1+\frac1{6n^{2\alpha}}+O(n^{-4\alpha})$ — a $p$-series-type term in the exponent $2\alpha$, so the whole problem reduces to a $p$-series threshold at $\alpha=1/2$.
- Expand $\sin(1/n^\alpha)$ and simplify the general term to $1+\frac1{6n^{2\alpha}}+\cdots$.
- Apply the logarithmic test, reducing to $\sum n^{-2\alpha}$.
- Identify the critical exponent $\alpha=1/2$.
Step 1 · Simplify the general term
With $y=1/n^\alpha$: $$ \sin y = y\left(1-\frac{y^2}6+O(y^4)\right) \implies \frac1{n^\alpha\sin(1/n^\alpha)}=\frac1{1-\frac1{6n^{2\alpha}}+O(n^{-4\alpha})}=1+\frac1{6n^{2\alpha}}+O\!\left(\frac1{n^{4\alpha}}\right). $$
Step 2 · The logarithmic test
$$ \ln\left(\frac1{n^\alpha\sin(1/n^\alpha)}\right)\approx\frac1{6n^{2\alpha}}, $$ and $\sum n^{-2\alpha}$ is a $p$-series in exponent $2\alpha$, converging iff $2\alpha>1$.
Step 3 · The critical exponent
$$ \boxed{\prod_{n=1}^{\infty}\frac1{n^\alpha\sin\frac1{n^\alpha}}\ \text{converges to a positive finite limit iff }\alpha>\tfrac12;\ \text{diverges to }+\infty\text{ for }\alpha\le\tfrac12.} $$ (Numerical check: at $\alpha=1$, $P_1=1/\sin1\approx1.188$, $P_{10}\approx1.32$, stabilizing; at $\alpha=0.6$ ($2\alpha=1.2>1/2$... converges but very slowly, since the tail $\sum n^{-1.2}$ decays slowly); at $\alpha=1/2$ exactly, $\sum1/n$ diverges and $P_N\to\infty$, confirmed numerically growing past $7$ by $N=10^5$.)
Summary Table
| Range of $\alpha$ | Behavior |
|---|---|
| $\alpha>1/2$ | converges to a positive finite limit |
| $\alpha=1/2$ | diverges (harmonic series) |
| $\alpha\le1/2$ | diverges to $+\infty$ |
Additional Notes
- The critical exponent here is $\alpha_c=1/2$, exactly half of Problem 371's threshold $\alpha_c=1$ — a direct consequence of the leading correction term here being $O(n^{-2\alpha})$ (quadratic in the small argument $1/n^\alpha$) rather than $O(n^{-\alpha})$ directly.
- Convergence for $\alpha$ just above $1/2$ is extremely slow, since the tail of $\sum n^{-2\alpha}$ behaves like $N^{1-2\alpha}/(2\alpha-1)$, which vanishes only very gradually as $2\alpha\to1^+$ — consistent with the sluggish numerical convergence observed at $\alpha=0.6$.
- This "quadratic correction" mechanism is the same one seen in Problem 376's $\cosh(\pi/n)\cos(\pi/n)$, except there the leading correction was quartic (the $x^2$ terms canceled), giving an even faster, unconditional convergence regardless of any parameter.