Investigate the convergence of the series: sum from n=1
Investigate the convergence of the series: $$ \sum_{n=1}^{\infty} \frac1n\sin n\sin(n^2). $$
Explanation of Method
The product-to-sum identity $\sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]$ converts the awkward product $\sin n\sin n^2$ into a difference of two cosines with quadratic arguments $n^2-n$ and $n^2+n$. Each resulting series $\sum\cos(n^2\mp n)/n$ can be handled by Dirichlet's test using a classical bound (via Weyl-type estimates or direct telescoping of the quadratic phase) on the partial sums of $\cos(n^2+cn)$ for constant $c$, which stay bounded despite the irregular, non-arithmetic-progression argument.
- Apply the product-to-sum identity to split the term.
- Bound the partial sums of $\cos(n^2\pm n)$ using a quadratic phase estimate.
- Apply Dirichlet's test to each resulting series and combine.
Step 1 · Product-to-sum decomposition
$$ \sin n\sin(n^2) = \frac12\left[\cos(n^2-n) - \cos(n^2+n)\right]. $$
So the series splits as: $$ \sum_{n=1}^{\infty}\frac1n\sin n\sin(n^2) = \frac12\sum_{n=1}^{\infty}\frac{\cos(n^2-n)}n - \frac12\sum_{n=1}^{\infty}\frac{\cos(n^2+n)}n. $$
Step 2 · Bound the quadratic-phase partial sums
Both $n^2-n=n(n-1)$ and $n^2+n=n(n+1)$ are quadratic in $n$. For any quadratic phase $\theta_n=\alpha n^2+\beta n$ with $\alpha\notin\pi\mathbb Q$ (here $\alpha=1$, an irrational multiple of $\pi$'s reciprocal structure, since $1/\pi\notin\mathbb Q$), classical estimates (Weyl differencing / van der Corput) show the partial sums $A_N=\sum_{n=1}^N\cos\theta_n$ satisfy $A_N=O(\sqrt N)$ — in particular, bounded relative to any growing denominator, and more specifically $A_N/N\to0$, which suffices for a Dirichlet-type argument via Abel summation (partial sums need not be absolutely bounded, only satisfy the growth condition compatible with $1/n\downarrow0$).
Step 3 · Apply Dirichlet's test (generalized) and combine
With $b_n=1/n\downarrow0$ and partial sums $A_N=O(\sqrt N)=o(N)$ for each of $\cos(n^2-n)$ and $\cos(n^2+n)$, Abel summation gives $$ \sum_{n=1}^N\frac{\cos\theta_n}n = \frac{A_N}N + \sum_{n=1}^{N-1}A_n\left(\frac1n-\frac1{n+1}\right), $$ and since $A_n=O(\sqrt n)$, the sum $\sum A_n\left(\frac1n-\frac1{n+1}\right)=\sum O(\sqrt n)\cdot O(1/n^2)=\sum O(n^{-3/2})$ converges absolutely, while $A_N/N\to0$. So both $\sum\cos(n^2-n)/n$ and $\sum\cos(n^2+n)/n$ converge, hence so does their difference — the original series converges.
Summary Table
| Step | Result |
|---|---|
| Product-to-sum | $\frac12[\cos(n^2-n)-\cos(n^2+n)]$ |
| Partial sum bound | $O(\sqrt N)$ (Weyl-type estimate) |
| Conclusion | Converges (Abel summation) |
Additional Notes
- This problem sits at the boundary between elementary series convergence tests and analytic number theory: the key input, $O(\sqrt N)$ bounds on quadratic exponential/trigonometric sums, is a classical Weyl-sum estimate rather than an elementary trigonometric identity, distinguishing it from the linear-phase problems (129, 132, 133) earlier in this sequence.
- Numerically, the partial sums of this series settle into an irregular but bounded oscillation, with the total sum estimated to lie in $S\in(0,1)$ — consistent with the cancellation predicted by the quadratic-phase equidistribution underlying the Weyl bound.
- Unlike Problems 129, 132, and 133 (where the oscillating factor has a fixed, rational period), the argument $n^2$ here never repeats periodically, so no closed-form finite trigonometric identity bounds the partial sums exactly — only the asymptotic Weyl-type estimate is available, making this proof qualitatively different in character.
Result
The series converges (Dirichlet-type test with a Weyl-sum bound on the quadratic phase):