Using the Cauchy criterion, prove that the following
Using the Cauchy criterion, prove that the following series converges: $$ \sum_{n=1}^{\infty} \frac{\sin(x^{2n})}{n^2}. $$
Explanation of Method
No matter how the argument $x^{2n}$ behaves, $|\sin(\cdot)|$ is always capped at $1$, so each term is bounded in absolute value by $1/n^2$ — regardless of $x$. Since $\sum 1/n^2$ is a convergent $p$-series, its Cauchy tail sums can be made arbitrarily small, and that same bound transfers directly to the tail sums of the original series, for every $x$ simultaneously.
- Bound each term using $|\sin(\theta)|\le1$.
- Compare the resulting tail sum to the tail of the convergent series $\sum 1/n^2$.
- Apply the Cauchy criterion, noting the bound is uniform in $x$.
Step 1 · Bound the tail sum
For $m>n\ge1$, by the triangle inequality and $|\sin(x^{2k})|\le1$ for all $k,x$:
$$ \left|\sum_{k=n+1}^{m}\frac{\sin(x^{2k})}{k^2}\right| \le \sum_{k=n+1}^{m}\frac{|\sin(x^{2k})|}{k^2} \le \sum_{k=n+1}^{m}\frac{1}{k^2}. $$
Step 2 · Use convergence of the dominating $p$-series
$\sum_{k=1}^{\infty}\frac1{k^2}$ is a convergent $p$-series ($p=2>1$, equal to $\zeta(2)=\pi^2/6$). By the Cauchy criterion applied to this series, for every $\varepsilon>0$ there exists $N$ such that for all $m>n\ge N$:
$$ \sum_{k=n+1}^{m}\frac{1}{k^2} < \varepsilon. $$
Step 3 · Apply the Cauchy criterion to the original series
Combining the two steps, for all $m>n\ge N$:
$$ \left|\sum_{k=n+1}^{m}\frac{\sin(x^{2k})}{k^2}\right| \le \sum_{k=n+1}^{m}\frac{1}{k^2} < \varepsilon. $$
This is the Cauchy criterion for $\sum_{n=1}^{\infty}\frac{\sin(x^{2n})}{n^2}$, and it holds with the same $N$ regardless of $x$, so the series converges for every $x\in\mathbb R$, absolutely and uniformly in $x$.
Summary Table
| Quantity | Bound |
|---|---|
| $|\sin(x^{2n})/n^2|$ | $\le 1/n^2$, for every $x$ |
| Dominating series $\sum 1/n^2$ | Convergent $p$-series ($p=2$) |
| Conclusion | Converges absolutely, uniformly in $x$ |
Additional Notes
- This is the Weierstrass M-test in disguise: since $|\sin(x^{2n})/n^2| \le M_n := 1/n^2$ for every $x$, and $\sum M_n$ converges, the series converges absolutely and uniformly on all of $\mathbb R$.
- Uniform convergence matters because the same $N$ works for every value of $x$ — no matter how wildly $x^{2n}$ behaves for different $x$, the bound $1/n^2$ never depends on it.
- The specific form of the argument, $x^{2n}$, is irrelevant to the convergence proof; any bounded function of $n$ and $x$ (not just $\sin$) composed with a $1/n^2$ weight would converge by the identical argument.
Result
The series converges absolutely and uniformly for all $x\in\mathbb{R}$, by the Cauchy criterion: