Using the Weierstrass test, show that the series sum
Using the Weierstrass test, show that the series $\displaystyle\sum_{n=1}^{\infty}\frac{\sin nx}{n!}$ converges uniformly on $(-\infty,\infty)$.
Explanation of Method
Bounding $|\sin nx|\le1$ dominates the series by $\sum1/n!$, which converges (it's essentially the series for $e$). The M-test applies immediately, and a closed form via Euler's formula shows the sum is the elegant bounded function $e^{\cos x}\sin(\sin x)$.
- Bound each term by $1/n!$, uniformly in $x$.
- Confirm $\sum1/n!=e-1$ converges.
- Apply the M-test, and identify the closed form via Euler's formula.
Step 1 · Uniform bound
$|\sin nx|\le1$ for all $x,n$, so $\left|\dfrac{\sin nx}{n!}\right|\le\dfrac1{n!}$.
Step 2 · Majorant series
$\sum_{n=1}^{\infty}1/n!=e-1<\infty$.
Step 3 · M-test and closed form
By the Weierstrass M-test with $M_n=1/n!$, the series converges uniformly and absolutely on $\mathbb R$. Using $\sin(nx)=\mathrm{Im}(e^{inx})$ and $\sum_{n=0}^{\infty}z^n/n!=e^z$: $$ \sum_{n=1}^{\infty}\frac{\sin nx}{n!} = \mathrm{Im}\left(e^{e^{ix}}-1\right) = \mathrm{Im}\left(e^{\cos x}e^{i\sin x}\right) = e^{\cos x}\sin(\sin x). $$
Summary Table
| Quantity | Value |
|---|---|
| Majorant | $M_n=1/n!$, $\sum M_n=e-1$ |
| Closed form | $e^{\cos x}\sin(\sin x)$ |
Additional Notes
- This is the fastest-converging series in the $\sum(\text{trig})/n^{\text{something}}$ family in this batch, since factorial decay dominates polynomial decay (like $n^a$ in Problem 192) enormously — uniform convergence holds with no restriction on any parameter.
- The closed form $e^{\cos x}\sin(\sin x)$ is manifestly a bounded, smooth, $2\pi$-periodic function, a nice concrete confirmation that the sum inherits continuity (and indeed smoothness) directly from uniform convergence of a series of smooth terms.
- The same complex-exponential technique used here (writing $\sin nx=\mathrm{Im}(e^{inx})$ and summing the resulting complex geometric-type series) is the identical trick used to evaluate Problem 142's trigonometric double series in closed form.
Result
Converges uniformly and absolutely on $\mathbb R$: