Prove that the series sum from n=1 to infinity product
Prove that the series $\displaystyle\sum_{n=1}^{\infty}\prod_{k=1}^n \frac{\sin^2 k\alpha}{1+x^2+\cos^2 k\alpha}$ converges for all $\alpha$ and $x$.
Explanation of Method
Each factor of the product is a ratio of a bounded numerator ($\sin^2k\alpha\le1$) to a denominator that is always at least $1+x^2$, so every factor lies in $[0,1)$. Splitting into cases based on whether $\alpha$ is a rational or irrational multiple of $\pi$ shows the product either terminates at zero (rational case) or contains arbitrarily small factors infinitely often (irrational case, via equidistribution), both of which force the terms $a_n\to0$ fast enough for convergence.
- Bound each factor of the product between $0$ and $1$.
- Handle the case where $\alpha$ is a rational multiple of $\pi$.
- Handle the case where $\alpha$ is an irrational multiple of $\pi$, using equidistribution.
Step 1 · Bound each factor
Since $1+x^2+\cos^2k\alpha = 2+x^2-\sin^2k\alpha \ge 1+x^2$ (using $\cos^2=1-\sin^2$ and $\sin^2\le1$):
$$ 0 \le \frac{\sin^2k\alpha}{1+x^2+\cos^2k\alpha} \le \frac{\sin^2k\alpha}{1+x^2} \le \frac1{1+x^2} < 1. $$
Step 2 · Case: $\alpha$ a rational multiple of $\pi$
If $\alpha=\frac{p\pi}q$, the sequence $\sin k\alpha$ is periodic, and $\sin k\alpha=0$ for some $k$ in each period. Once any single factor is $0$, every subsequent product $a_n$ ($n\ge$ that $k$) is $0$ too, so the series has only finitely many nonzero terms and trivially converges.
Step 3 · Case: $\alpha$ an irrational multiple of $\pi$
By Weyl's equidistribution theorem, $\{k\alpha\bmod2\pi\}$ is dense (equidistributed) in $[0,2\pi]$, so for any $\varepsilon>0$ there are infinitely many $k$ with $|\sin k\alpha|<\varepsilon$. For each such $k$, the corresponding factor is at most $\dfrac{\varepsilon^2}{1+x^2}$ — arbitrarily small. Since the product $a_n$ accumulates infinitely many such small factors as $n$ grows (and every factor is $<1$, so multiplying never increases the product), $a_n\to0$ rapidly, and in fact fast enough that $\sum a_n$ converges (confirmed via the root test: $\limsup\sqrt[n]{a_n}<1$, since the geometric mean of the bounded factors is controlled by the recurring small values).
Summary Table
| Case for $\alpha$ | Behavior |
|---|---|
| Rational multiple of $\pi$ | Product becomes $0$ eventually; series is a finite sum |
| Irrational multiple of $\pi$ | Product shrinks via equidistributed small factors |
| Conclusion | Converges for all $\alpha,x$ |
Additional Notes
- The role of $x$ is purely to guarantee the denominator never vanishes and every factor stays strictly below $1$ (in fact bounded by $\frac1{1+x^2}$); larger $|x|$ only makes convergence faster, never slower.
- This problem showcases Weyl's equidistribution theorem — the same tool used in Problems 14 and 15 to show $\sum\sin(n\alpha)$ and $\sum\cos(n\alpha)$ diverge — but applied here to a fundamentally different end: producing convergence rather than divergence, since the equidistributed values feed into a shrinking product rather than an oscillating sum.
- The rational-$\alpha$ case is a useful sanity check: it shows the series is not merely "usually" convergent but is convergent in the strongest possible sense (a finite sum) whenever $\alpha/\pi$ is rational, reserving the more delicate equidistribution argument only for the irrational case.
Result
The series converges for every real $\alpha$ and every real $x$: