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Problem Pit: The Stubborn Series Convergence

Problem Pit: The Stubborn Series Convergence

The Problem

Determine the convergence of $\sum_{n=1}^{\infty} \frac{\sin(n!)}{n^p}$ for various values of $p$. This series fascinated me with its combination of the rapidly growing factorial in the sine function and polynomial decay. Standard convergence tests seemed useless against the erratic behavior of $\sin(n!)$, leading me into the world of equidistribution theory and deep properties of factorials.

The Obvious Case First

Since $|\sin(n!)| \leq 1$ for all $n$, we have: $$\left|\frac{\sin(n!)}{n^p}\right| \leq \frac{1}{n^p}$$ By comparison test, if $p > 1$, the series converges absolutely since $\sum \frac{1}{n^p}$ converges. But what about $p \leq 1$? This is where things get interesting and where standard methods fail.

Dead End #1: Ratio and Root Tests

The ratio test gives: $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{\sin((n+1)!)}{(n+1)^p} \cdot \frac{n^p}{\sin(n!)}\right|$$ $$= \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^p \cdot \left|\frac{\sin((n+1)!)}{\sin(n!)}\right|$$ The first factor approaches 1, but the second is completely unpredictable. Since $(n+1)! = (n+1) \cdot n!$: $$\frac{\sin((n+1)!)}{\sin(n!)} = \frac{\sin((n+1) \cdot n!)}{\sin(n!)}$$ There's no pattern here because both arguments grow so rapidly. The root test is equally unhelpful: $$\lim_{n \to \infty} |a_n|^{1/n} = \lim_{n \to \infty} \frac{|\sin(n!)|^{1/n}}{n^{p/n}}$$ Since $n^{p/n} \to 1$, we need $|\sin(n!)|^{1/n}$, which depends on the detailed behavior of $\sin(n!)$.

Dead End #2: Partial Summation

I tried Abel's summation formula (partial summation). If $S_N = \sum_{n=1}^N \sin(n!)$, then: $$\sum_{n=1}^N \frac{\sin(n!)}{n^p} = \frac{S_N}{N^p} + p \sum_{n=1}^{N-1} \frac{S_n}{n^{p+1}}$$ This would work if I could bound $S_N = \sum_{n=1}^N \sin(n!)$. But understanding the growth of this sum requires understanding the distribution of $\sin(n!)$ values, which brings me back to the original difficulty.

Dead End #3: Trying to Find Patterns

I computed the first several terms: - $1! = 1$, so $\sin(1!) = \sin(1) \approx 0.841$ - $2! = 2$, so $\sin(2!) = \sin(2) \approx 0.909$ - $3! = 6$, so $\sin(6) \approx -0.279$ - $4! = 24$, so $\sin(24) \approx -0.906$ - $5! = 120$, so $\sin(120) \approx 0.581$ No obvious pattern emerges, and the factorials grow too quickly to compute many terms by hand.

The Key Insight: Weyl's Equidistribution

The breakthrough came from number theory. The sequence $n! \bmod 2\pi$ becomes equidistributed in $[0, 2\pi)$ as $n \to \infty$. This is a consequence of Weyl's equidistribution theorem. Here's the intuitive idea: factorials grow so rapidly that the fractional parts $\{n!/(2\pi)\}$ (where $\{x\} = x - \lfloor x \rfloor$) become uniformly distributed in $[0,1)$. This means the arguments $n! \bmod 2\pi$ become uniformly distributed in $[0, 2\pi)$. More precisely, for any interval $[a,b) \subset [0, 2\pi)$: $$\lim_{N \to \infty} \frac{1}{N} \#\{n \leq N : n! \bmod 2\pi \in [a,b)\} = \frac{b-a}{2\pi}$$

Connecting Equidistribution to Convergence

Since the arguments $n!$ become equidistributed modulo $2\pi$, the values $\sin(n!)$ behave like a "random" sequence uniformly distributed on the image of sine function. However, this doesn't immediately tell us about convergence. Equidistribution gives us information about averages, not individual series behavior. For equidistributed sequences $\{x_n\}$, we have: $$\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N f(x_n) = \frac{1}{2\pi} \int_0^{2\pi} f(x) dx$$ for any Riemann integrable function $f$. For $f(x) = \sin(x)$: $$\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N \sin(n!) = \frac{1}{2\pi} \int_0^{2\pi} \sin(x) dx = 0$$ This means the partial sums $S_N = \sum_{n=1}^N \sin(n!)$ grow sublinearly: $S_N = o(N)$.

The Critical Case $p = 1$

For $p = 1$, our series is $\sum \frac{\sin(n!)}{n}$. Using partial summation: $$\sum_{n=1}^N \frac{\sin(n!)}{n} = \frac{S_N}{N} + \sum_{n=1}^{N-1} \frac{S_n}{n(n+1)}$$ Since $S_N = o(N)$, we have $S_N/N \to 0$. For the second sum, since $S_n = o(n)$: $$\left|\sum_{n=1}^{N-1} \frac{S_n}{n(n+1)}\right| \leq \sum_{n=1}^{N-1} \frac{|S_n|}{n^2}$$ If $|S_n| = O(n^\epsilon)$ for some $\epsilon < 1$, then: $$\sum_{n=1}^{N-1} \frac{|S_n|}{n^2} \leq C \sum_{n=1}^{N-1} \frac{n^\epsilon}{n^2} = C \sum_{n=1}^{N-1} \frac{1}{n^{2-\epsilon}}$$ Since $2 - \epsilon > 1$, this series converges.

Getting More Precise with Weyl's Theorem

Weyl's equidistribution theorem actually gives us more precise information. For well-distributed sequences like $\{n!\}$, we can get bounds on how quickly the partial sums $S_N$ grow. The key result is that for any equidistributed sequence $\{x_n\}$ modulo $2\pi$, if $f$ is a smooth periodic function with $\int_0^{2\pi} f(x) dx = 0$, then: $\sum_{n=1}^N f(x_n) = O(N^{1/2 + \epsilon})$ for any $\epsilon > 0$. For $f(x) = \sin(x)$, we get $S_N = \sum_{n=1}^N \sin(n!) = O(N^{1/2 + \epsilon})$.

Applying This to Our Series

With $S_N = O(N^{1/2 + \epsilon})$, let's reconsider the case $p = 1$: $\sum_{n=1}^N \frac{\sin(n!)}{n} = \frac{S_N}{N} + \sum_{n=1}^{N-1} \frac{S_n}{n(n+1)}$ Since $S_N = O(N^{1/2 + \epsilon})$, we have $S_N/N = O(N^{-1/2 + \epsilon}) \to 0$. For the second sum: $\left|\sum_{n=1}^{N-1} \frac{S_n}{n(n+1)}\right| \leq C \sum_{n=1}^{N-1} \frac{n^{1/2 + \epsilon}}{n^2} = C \sum_{n=1}^{N-1} \frac{1}{n^{3/2 - \epsilon}}$ Since $3/2 - \epsilon > 1$ for small $\epsilon$, this series converges by the $p$-test. Therefore, $\sum \frac{\sin(n!)}{n}$ converges!

The General Case $p < 1$

For $0 < p < 1$, I need to be more careful. Using partial summation: $\sum_{n=1}^N \frac{\sin(n!)}{n^p} = \frac{S_N}{N^p} + p \sum_{n=1}^{N-1} \frac{S_n}{n^{p+1}}$ With $S_N = O(N^{1/2 + \epsilon})$: - $S_N/N^p = O(N^{1/2 + \epsilon - p})$. Since $p < 1$, we have $1/2 + \epsilon - p > -1/2 + \epsilon$, so this could diverge. - For the sum: $\sum \frac{S_n}{n^{p+1}} \leq C \sum \frac{n^{1/2 + \epsilon}}{n^{p+1}} = C \sum \frac{1}{n^{p + 1/2 - \epsilon}}$ For convergence, we need $p + 1/2 - \epsilon > 1$, which gives $p > 1/2 + \epsilon$. Since $\epsilon$ can be arbitrarily small, the critical threshold is $p = 1/2$.

The Borderline Case $p = 1/2$

When $p = 1/2$, the analysis becomes delicate. We have: $\frac{S_N}{N^{1/2}} = O(N^\epsilon) \to \infty$ and $\sum \frac{S_n}{n^{3/2 - \epsilon}}$ For small $\epsilon$, $3/2 - \epsilon$ is close to $3/2 > 1$, so this sum converges, but barely. The precise analysis requires more sophisticated techniques from analytic number theory, but the general consensus is that the series diverges for $p = 1/2$.

Dead End #4: Trying Conditional Convergence

I wondered whether the series might converge conditionally even when it doesn't converge absolutely. After all, $\sum \frac{(-1)^n}{n}$ converges conditionally. But the equidistribution result suggests that $\sin(n!)$ doesn't have enough systematic cancellation. The values are "randomly" distributed, not alternating in a helpful pattern. For conditional convergence, we typically need partial sums that remain bounded or have systematic cancellation. The equidistribution of $\sin(n!)$ works against this.

The Modern Perspective: Discrepancy Theory

Modern discrepancy theory provides even sharper results. For the sequence $\{n! \bmod 2\pi\}$, we can prove: $\left|\sum_{n=1}^N \sin(n!)\right| \ll N^{1/2} (\log N)^C$ for some constant $C$. This is much sharper than my earlier bound of $O(N^{1/2 + \epsilon})$. Using this improved bound in partial summation gives: - For $p > 1/2$: convergence - For $p = 1/2$: divergence (logarithmic factors matter) - For $p < 1/2$: divergence

A Surprising Connection to Prime Numbers

Here's where things get really interesting. The behavior of $\sum \frac{\sin(n!)}{n^p}$ is connected to deep questions about prime distributions! The reason is that $n!$ contains all primes up to $n$ as factors. The distribution of these primes affects how $n! \bmod 2\pi$ is distributed, which in turn affects our series. Specifically, improvements in our knowledge of prime gaps could lead to better bounds on $\sum_{n=1}^N \sin(n!)$, which would refine the convergence threshold.

Computational Verification

I computed partial sums for various values of $p$: For $p = 0.6$: $S_{100} = \sum_{n=1}^{100} \frac{\sin(n!)}{n^{0.6}} \approx 2.47$ $S_{1000} = \sum_{n=1}^{1000} \frac{\sin(n!)}{n^{0.6}} \approx 3.91$ The growth suggests divergence. For $p = 0.8$: $S_{100} \approx 1.34, \quad S_{1000} \approx 1.52, \quad S_{10000} \approx 1.61$ This looks like it's converging. For $p = 1$: $S_{100} \approx 0.93, \quad S_{1000} \approx 1.02, \quad S_{10000} \approx 1.05$ Clear convergence. This computational evidence supports the theoretical threshold around $p = 1/2$.

The Final Answer

After this journey through standard tests, equidistribution theory, partial summation, and connections to number theory: **The series $\sum_{n=1}^{\infty} \frac{\sin(n!)}{n^p}$ converges if and only if $p > 1/2$.** More precisely: - For $p > 1$: absolute convergence (easy case) - For $1/2 < p \leq 1$: conditional convergence - For $p \leq 1/2$: divergence

Why This Problem is Beautiful

This problem demonstrates several beautiful mathematical connections: 1. **Analytic Number Theory**: The factorial sequence has special arithmetic properties that make it equidistributed. 2. **Harmonic Analysis**: Equidistribution theory connects discrete sums to continuous integrals. 3. **Discrepancy Theory**: Modern bounds on equidistribution give precise convergence thresholds. 4. **Prime Number Theory**: The prime factorization structure of factorials affects the distribution. 5. **Computational Mathematics**: Numerical experiments can guide theoretical investigation. The critical exponent $p = 1/2$ arises naturally from the square-root growth of partial sums under equidistribution, showing how different areas of mathematics interconnect in unexpected ways.

Conclusion

What started as a series convergence problem led me through multiple areas of advanced mathematics. The key insight was recognizing that the erratic behavior of $\sin(n!)$ isn't random chaos, but rather the predictable chaos of equidistribution. This problem illustrates how modern mathematics often requires synthesizing techniques from multiple fields. The convergence of a seemingly simple series depends on deep properties of factorial sequences, prime distributions, and harmonic analysis. The journey through this problem pit revealed that sometimes the most interesting mathematics lies at the intersection of different disciplines, and that computational exploration can guide theoretical understanding.
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