Investigate the uniform convergence of the series sum
Investigate the uniform convergence of the series $\displaystyle\sum_{n=1}^{\infty}\frac{nx}{1+n^4x^2}$ on $(-\infty,\infty)$.
Explanation of Method
The sharpest possible $x$-independent bound on each term is $1/(2n)$ (attained at $x=1/n^2$, via AM-GM), but $\sum1/(2n)$ diverges, so the Weierstrass M-test cannot certify uniform convergence. This is not just a failure of the test: testing the Cauchy criterion directly at the moving points $x_m=1/m^2$ shows the tail sums stay bounded away from $0$, proving uniform convergence genuinely fails on any interval containing $0$. Restricted to $\{|x|\ge\delta\}$, however, a sharper bound recovers a convergent majorant.
- Find the sharp $x$-independent bound $1/(2n)$ and note its majorant series diverges.
- Test the Cauchy criterion directly at $x_m=1/m^2$ to show non-uniform convergence on $\mathbb R$.
- Recover uniform convergence on $\{|x|\ge\delta\}$ via a sharper, $\delta$-dependent bound.
Step 1 · Sharp bound and its failure
By AM-GM, $1+n^4x^2\ge2n^2|x|$, so $\left|\dfrac{nx}{1+n^4x^2}\right|\le\dfrac1{2n}$, with equality at $x=1/n^2$. Since $\sum1/(2n)$ diverges, the Weierstrass M-test does not apply on all of $\mathbb R$.
Step 2 · Cauchy criterion fails at $x_m=1/m^2$
At $x_m=1/m^2$, writing $k=n/m$: $u_n(x_m)=\dfrac{n/m^2}{1+n^4/m^4}=\dfrac{g(k)}m$ with $g(k)=\dfrac k{1+k^4}$, decreasing on $[1,\infty)$ with $g(2)=2/17$. For $n$ ranging from $m$ to $2m$ ($m+1$ values, $k\in[1,2]$): $$ \sum_{n=m}^{2m}u_n(x_m) \ge (m+1)\cdot\frac{g(2)}m \ge \frac2{17} $$ for every $m$ — a tail sum bounded away from $0$ regardless of how large $m$ is. So the Cauchy criterion fails uniformly, and the series does not converge uniformly on $\mathbb R$ (nor on $(0,\infty)$, since $x_m\to0^+$).
Step 3 · Uniform convergence on $\{|x|\ge\delta\}$
For $|x|\ge\delta>0$: dropping the $1$ in the denominator, $\left|\dfrac{nx}{1+n^4x^2}\right|\le\dfrac{n|x|}{n^4x^2}=\dfrac1{n^3|x|}\le\dfrac1{n^3\delta}$. Since $\sum1/(n^3\delta)=\zeta(3)/\delta<\infty$, the Weierstrass M-test now applies: uniform convergence holds on $\{x:|x|\ge\delta\}$ for any fixed $\delta>0$.
Summary Table
| Domain | Uniform convergence? |
|---|---|
| $\mathbb R$ (or any interval containing $0$) | No |
| $\{x:|x|\ge\delta\}$, any $\delta>0$ | Yes |
Additional Notes
- This is a genuine "moving bump" phenomenon for a series (rather than a single sequence, as in Problem 177): each term's individual peak, located at $x=1/n^2$, contributes a fixed amount to a nearby tail sum, and since these peaks march toward $0$ as $n$ grows, the series' convergence near $x=0$ is permanently disrupted.
- This problem is the standard textbook counterexample illustrating that failure of the Weierstrass M-test does not by itself prove non-uniform convergence (the test is only sufficient, not necessary) — one must fall back on the Cauchy criterion directly, as done in Step 2, to settle the question rigorously.
- The contrast between Steps 1 and 3 — the same series failing near $x=0$ but succeeding once bounded away from it — echoes the same qualitative pattern seen in Problems 176, 181, and 182, where uniform convergence holds on any closed region excluding a single troublesome boundary point.
Result
Not uniformly convergent on $\mathbb R$, but uniformly convergent away from $x=0$: