Can the series sum from n=1 to infinity arctan x/(n²), x
Can the series $\displaystyle\sum_{n=1}^{\infty}\arctan\frac{x}{n^2}$, $x\in(-\infty,\infty)$, be differentiated term by term?
Explanation of Method
Each term is bounded via $|\arctan t|\le|t|$, giving uniform convergence on every bounded interval. The formal derivative $u_n'(x)=n^2/(n^4+x^2)$ satisfies the clean $x$-independent bound $1/n^2$, so the Weierstrass M-test applies to the derivative series on all of $\mathbb R$ — both hypotheses of the differentiation theorem hold globally.
- Bound each term to establish convergence on bounded intervals.
- Compute the formal derivative $u_n'(x)=n^2/(n^4+x^2)$.
- Apply the Weierstrass M-test to the derivative series on all of $\mathbb R$.
Step 1 · Convergence of the original series
For $x\in[-a,a]$: $\left|\arctan\dfrac x{n^2}\right|\le\arctan\dfrac a{n^2}\le\dfrac a{n^2}$ (using $|\arctan t|\le|t|$). Since $\sum a/n^2=a\pi^2/6<\infty$, the Weierstrass M-test gives uniform convergence on $[-a,a]$ for every $a>0$, hence pointwise convergence on all of $\mathbb R$.
Step 2 · The formal derivative
$$ u_n'(x) = \frac{d}{dx}\left[\arctan\frac x{n^2}\right] = \frac1{1+(x/n^2)^2}\cdot\frac1{n^2} = \frac{n^2}{n^4+x^2}. $$
Step 3 · Uniform convergence of the derivative series
For every $x\in\mathbb R$: $$ \frac{n^2}{n^4+x^2} \le \frac{n^2}{n^4} = \frac1{n^2}. $$ Since $\sum1/n^2=\pi^2/6<\infty$, the Weierstrass M-test gives uniform (and absolute) convergence of $\sum n^2/(n^4+x^2)$ on all of $\mathbb R$ — no restriction to a bounded interval is needed. Since both hypotheses of the differentiation theorem hold globally, term-by-term differentiation is valid on all of $\mathbb R$: $$ \frac{d}{dx}\left[\sum_{n=1}^{\infty}\arctan\frac x{n^2}\right] = \sum_{n=1}^{\infty}\frac{n^2}{n^4+x^2}. $$
Summary Table
| Series | Behavior |
|---|---|
| $\sum\arctan(x/n^2)$ | Uniformly convergent on every $[-a,a]$ |
| $\sum n^2/(n^4+x^2)$ (derivative) | Uniformly convergent on all of $\mathbb R$ ($M_n=1/n^2$) |
| Term-by-term differentiation | Valid on $\mathbb R$ |
Additional Notes
- This is the strongest possible result among the three differentiation problems in this batch (216–218): here uniform convergence of the derivative series holds on the entire real line at once, not just a restricted subinterval as in Problem 217's Dirichlet-test argument.
- The bound $n^2/(n^4+x^2)\le1/n^2$ is structurally identical to the bounds used throughout Problems 194, 197, and 208 (drop a non-negative additive term from the denominator) — the same trick that repeatedly delivers global uniform convergence in this problem collection.
- Since $\arctan$ is smooth and bounded, and both the original series and its derivative series converge uniformly everywhere, the sum $f(x)=\sum\arctan(x/n^2)$ is not just differentiable but has a derivative given by a nicely convergent series — a much cleaner outcome than the trigonometric-series cases in Problems 216 and 217.
Result
Yes — valid on all of $\mathbb R$: