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I have the following exercise (not homework, I am just practicing in solving series problems): $$f(x) = \sum_{n=1}^{\infty} \frac{x}{n+x^2} \sin{\frac{x}{n}}$$ Determine whether the series converges pointwise or uniformly on:

  • [0,1]
  • [1,$\infty$)

My solution for the first case is based upon the fact that $\sin{\frac{x}{n}} < \frac{x}{n}$ for positive $x$, hence we can obtain the following bound: $$\left|\frac{x}{n+x^2} \sin{\frac{x}{n}}\right| \le \left|\frac{x^2}{n^2 + nx^2}\right| \le \left|\frac{1}{n^2}\right|$$ for $x \in [0,1]$, thus by Weierstrass M-test the series converges uniformly on this interval.

For the second interval $[0, \infty)$, using similar reasoning I can show the pointwise convergence, but I don't know how to prove or disprove the uniform convergence. The necessary condition for the uniform convergence seems to be satisfied, but that is all I've got. Any help would be much appreciated.

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The series does not converge uniformly on $[0,\infty)$. I will show that it does not satisfy the uniform Cauchy criterion.

Choose $N\in\mathbb{N}$ and $x=N$. Then $$ \min\{\sin\frac{x}{n}:N\le n\le2\,N\}=\sin\frac12>0. $$ Now, if $x=N$, $$ \sum_{n=N}^{2N}\frac{x}{n+x^2}\sin\frac{x}{n}\ge\sin\frac12\sum_{n=N}^{2N}\frac{N}{n+N^2}\ge (N+1)\frac{N}{2\,N+N^2}\sin\frac12>\frac12\sin\frac12>0. $$

Let $\epsilon=(1/2)\sin(1/2)>0$. If the series were uniformly convergent on $[1,\infty)$, by Cauchy's uniform criterion, there would exist $N_\epsilon$ such that if $N_\epsilon\le m<n$ then $$ \Bigl|\sum_{k=m}^n\frac{x}{n+x^2}\sin\frac{x}{n}\Bigr|\le\epsilon\quad\forall x\ge1. $$ Taking $m=N>N_\epsilon$, $n=2\,N$ and $x=N$, we arive at a conradiction.

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  • $\begingroup$ Did you mean that the series does not converge uniformly on [1, $\infty$)? Because as you wrote, $N \in \mathbb{N}$ and since in your proof $x=N$, it belongs to $[1,\infty)$? Am I right? $\endgroup$ – Pukki Feb 13 '14 at 16:13
  • $\begingroup$ It does not converge uniformly on $[0,\infty)$, and it does not converge uniformly on $[1,\infty)$. $\endgroup$ – Julián Aguirre Feb 13 '14 at 21:40
  • $\begingroup$ Sorry, but I can't see how your proof shows that the series does not converge on [0,$\infty$). You are picking $x = N$, where $N \in \mathbb{N}$, so it can't be less than 1. Moreover, if x can be 0 then $\min\{\sin{\frac{x}{n}}: N \le n \le 2N\} = 0$. And, after all, I can't find a mistake in my proof of the fact that this series converges uniformly on $[0,1]$, could you please be so kind as to explain me this? $\endgroup$ – Pukki Feb 14 '14 at 0:04
  • $\begingroup$ The series converges pointwise. I prove that the convergence is not uniform on $[1,\infty)$ because Cauchy's uniform criteria is not satisfied. See my edited answer. $\endgroup$ – Julián Aguirre Feb 14 '14 at 10:40

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