Can the series sum from n=1 to infinity (sin nx)/(n sqrt
Can the series $\displaystyle\sum_{n=1}^{\infty}\frac{\sin nx}{n\sqrt n}$, $x\in(-\infty,\infty)$, be integrated term by term?
Explanation of Method
Writing the coefficient as $n^{-3/2}$, the series is dominated by the convergent $p$-series $\sum n^{-3/2}$ ($p=3/2>1$), giving uniform convergence on $\mathbb R$ via the Weierstrass M-test. Combined with continuity of each term, term-by-term integration is valid on any interval.
- Bound each term by $1/n^{3/2}$ and apply the Weierstrass M-test.
- Confirm continuity of each term.
- Integrate term by term and verify the resulting series converges.
Step 1 · Uniform convergence
Since $|\sin nx|\le1$: $\left|\dfrac{\sin nx}{n^{3/2}}\right|\le\dfrac1{n^{3/2}}$. Since $\sum1/n^{3/2}$ is a convergent $p$-series ($p=3/2>1$), the Weierstrass M-test gives uniform (and absolute) convergence on $\mathbb R$.
Step 2 · Continuity
Each $u_n(x)=\sin(nx)/n^{3/2}$ is continuous on $\mathbb R$, hence integrable on any $[a,b]$.
Step 3 · Term-by-term integration
Since $\displaystyle\int\frac{\sin nx}{n^{3/2}}\,dx = -\frac{\cos nx}{n^{5/2}}+C$, term-by-term integration gives $$ \int_a^b\sum_{n=1}^{\infty}\frac{\sin nx}{n^{3/2}}\,dx = \sum_{n=1}^{\infty}\left[-\frac{\cos nx}{n^{5/2}}\right]_a^b = \sum_{n=1}^{\infty}\frac{\cos na-\cos nb}{n^{5/2}}. $$ Since $|\cos na-\cos nb|\le2$, this integrated series is dominated by $\sum2/n^{5/2}$, which converges ($p=5/2>1$), confirming the result is well-defined.
Summary Table
| Condition | Status |
|---|---|
| Uniform convergence on $\mathbb R$ | Yes ($M_n=1/n^{3/2}$) |
| Continuity of each term | Yes |
| Term-by-term integration | Valid on any $[a,b]$ |
Additional Notes
- This closes out the batch's pair of integration problems (219–220) alongside its pair of differentiation problems (216–218 minus one) with the same moral: integration only ever demands uniform convergence of the original series, never anything extra about a derivative series, making it structurally the easier of the two operations to justify.
- The decay rate $n^{-3/2}$ here sits between Problem 217's $n^{-2}$ (used for differentiation, where a stronger rate was needed to keep the derivative series convergent) and Problem 216's plain $n^{-1}$ (too weak even for the original derivative); for integration alone, $n^{-3/2}>1$ suffices comfortably.
- Each integration step trades a factor of $n$ for extra decay in the denominator (from $n^{-3/2}$ to $n^{-5/2}$), the reverse of how differentiation costs a factor of $n$ — the same trade-off responsible for Problem 217 needing $\sin(nx)/n^2$ rather than $\sin(nx)/n$ to survive one differentiation.
Result
Yes — valid on any interval $[a,b]$: