Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=1}^{\infty} \sin\frac{x}{n}, \qquad x\neq0. $$
Explanation of Method
Since $\sin t\sim t$ for small $t$, and $\frac xn\to0$ as $n\to\infty$, the general term $\sin\frac xn$ behaves like $\frac xn$ — a nonzero constant multiple of the harmonic term. Because $x\ne0$, that constant is nonzero, so the limit comparison test against $\sum1/n$ forces divergence, regardless of how small $|x|$ is.
- Expand $\sin\frac xn$ via its Taylor series.
- Apply the limit comparison test against $\sum1/n$, using the standard limit $\lim_{t\to0}\frac{\sin t}t=1$.
- Confirm by splitting the Taylor series into a divergent leading part and a convergent remainder.
Step 1 · Taylor expand
$$ \sin\frac xn = \frac xn - \frac{x^3}{6n^3} + O\!\left(\frac1{n^5}\right) \sim \frac xn \quad\text{for large } n. $$
Step 2 · Limit comparison test
With $b_n=\dfrac1n$, letting $h=\frac1n\to0^+$ and using $\lim_{u\to0}\frac{\sin u}u=1$:
$$ \lim_{n\to\infty}\frac{\sin(x/n)}{1/n} = \lim_{h\to0^+}\frac{\sin(xh)}{h} = \lim_{h\to0^+}\frac{\sin(xh)}{xh}\cdot x = x. $$
Since $x\neq0$, this limit is nonzero (in absolute value, $|x|>0$ and finite), so $\sum\left|\sin\frac xn\right|$ and $\sum\frac1n$ share the same convergence behavior. Since $\sum\frac1n$ diverges, $\sum\sin\frac xn$ diverges too.
Step 3 · Confirm by splitting the series
$$ \sum_{n=1}^{\infty}\sin\frac xn = \sum_{n=1}^{\infty}\left[\frac xn - \frac{x^3}{6n^3}+O\!\left(\frac1{n^5}\right)\right]. $$
The first part, $x\sum\frac1n$, diverges; the remaining parts (powers $\ge3$) converge. A divergent series plus a convergent one is divergent, confirming the conclusion.
Summary Table
| Quantity | Value |
|---|---|
| $\sin(x/n)$ asymptotics | $\sim x/n$ |
| Limit comparison ratio (vs. $1/n$) | $x\ne0$ |
| Effective exponent | $1$ (harmonic boundary) |
| Conclusion | Diverges for every $x\ne0$ |
Additional Notes
- The condition $x\ne0$ is essential: at $x=0$ every term is $\sin0=0$, and the series trivially converges to $0$ — a degenerate boundary case excluded from the problem statement.
- Even though $\sin\frac xn$ decreases to $0$ (a necessary condition for convergence), the decay rate matches the harmonic series exactly, which is the textbook counterexample showing $a_n\to0$ is not sufficient for convergence.
- This result holds regardless of how small $|x|$ is — even $x=0.0001$ produces a divergent series, since divergence depends only on the exponent of decay ($1/n$), not on the size of the constant multiplying it.
Result
The series diverges for every $x\neq0$: