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Let $$S_a(x)=\sum^{\infty }_{n=1}\frac{x}{n^a({1+nx^2})}$$ for $\ x\in \mathbb{R}$.

  1. I know $S_a(x)$ is pointwise convergent if $a>0$.
  2. I know $S_a(x)$ is uniformly convergent if $a>1/2$ (by the Weierstrass $M$-test).

But I can't proof $S_a(x)$ only converges pointwise if $0<a\le1/2$.

This means that if $0<a\le1/2$ , then $S_a(x)$ is not uniformly convergent.

I thought this was obvious, but I failed to prove it rigorously.

Is there anyone who can provide a rigorous and solid proof, other than an intuitive explanation?

Lucenaposition
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김석훈
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  • The fact that you could not prove uniform convergence does not mean that it is incorrect. First of all, check the necessary condition for uniform convergence. – zkutch Jun 04 '25 at 14:33
  • Consider $S_{2n}-S_n$ where $S_k$ is the partial sum and evaluate at $n^{-1/2}.$ Should be bounded away from $0$ for $a\le 1/2.$ – Ryszard Szwarc Jun 04 '25 at 14:34

2 Answers2

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The set where the uniform convergence is supposed to take place is not specified, I assume that it is $[0,1]$. We can show that $$ A_N:=\sup_{x\in [0,1]}\sum_{n=2^{N}+1}^{2^{N+1}}\frac{x}{n^a(1+nx^2)} $$ does not converge to $0$. Indeed, using that $n\leqslant 2^{N+1}$, a lower bound is $$ A_N\geqslant \frac 1{2^{(N+1)a}}\sup_{x\in [0,1]}\frac{x}{ (1+2^{N+1}x^2)}. $$ Then notice that the map $x\mapsto \frac{x}{ (1+2^{N+1}x^2)}$ reaches its maximum at $2^{-(N+1)/2}$ and we thus get $$ A_N\geqslant \frac{2^{(N+1)/2}}{2\times 2^{(N+1)a}}\geqslant\frac 12. $$

Davide Giraudo
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  • It's given in the question that $x\in \mathbb{R} $ – Mathson Jun 05 '25 at 04:37
  • @Mathson This is the definition of the summand in $S_a$. But nothing prevents from considering the uniform convergence on other sets than the whole real line. As a matter of fact, the convergence is uniform on $[1,+\infty)$ so there is a subtlety here. – Davide Giraudo Jun 05 '25 at 17:45
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The functions $f_n(x)=x/n^a(1+nx^2)$ form a decreasing sequence. The function $x/(1+x^2)$ is decreasing on $[1,\infty).$ Therefore the function $f_n(x)$ is decreasing on $[n^{-1/2},\infty)$. For $0<x\le 1$ there is $N$ such that $(2N)^{-1/2}\le x\le N^{-1/2}.$ Then, making use of the properties described above, we get $$ g(x):=\sum_{n=1}^\infty f_n(x)\ge \sum_{n=N+1}^{2N} f_n(x)\\ \ge Nf_{2N}(x) \ge Nf_{2N}(N^{-1/2})\\ =N{N^{-1/2}\over (2N)^a(1+2)}={1\over 3\cdot 2^a}N^{1/2-a}\ge {1\over 3\cdot 2^a} $$ Thus $$\inf_{0<x\le 1}g(x)>0=g(0)$$ which implies that $g(x)$ is not continuous at $0.$ Therefore the series cannot be convergent uniformly on $[0,1].$

Moreover for $a<1/2$ we get $$\lim_{x\to 0^+}g(x)=\infty $$

Ryszard Szwarc
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