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Assume the following sum: $$ \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{x^2+n^2}}-\frac{1}{n}\right), $$ where $x \in \mathbb{R}$. The problem is to determine whether the sum converges uniformly, i.e. whether there exists $s(x)$ such as $\sum f_n(x) \rightrightarrows s(x)$. My approach is following: I show $f_n \rightarrow 0$ for every $x$, as $$ f_n \sim \frac{1}{n^3}, \lim_{n \rightarrow \infty} \frac{f_n}{n^{-3}} = -\frac{x^2}{2}. $$ Then I can show $f_n \rightrightarrows 0$, as $$ |f_n(x)| \leq \frac{1}{n}, \forall x, $$ thus the sequence $f_n$ converges uniformly to zero, and here I got stuck. Is my approach correct? Could you help me find out on which interval the sum converges uniformly? Is the convergence absolute/local? What is the prescription of $s(x)$?

Thank you for any advice.

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    $\begingroup$ It is probably easier to deal with $F(x)$ written this way: $$ F(x)=-x^2\sum_{n\geq 1}\frac{1}{\sqrt{n^2+x^2}\left(n+\sqrt{n^2+x^2}\right)}$$ $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 18:26
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    $\begingroup$ Then you may notice that the $n$-th term of the series is bounded by $\frac{1}{2n^2}$, and $\left\{\frac{1}{2n^2}\right\}_{n\geq 1}$ is a summable sequence, hence we have uniform convergence over any compact subset of the real line. $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 18:29
  • $\begingroup$ Took me a while to get it, now I see: $f_n = \frac{n-\sqrt{n^2+x^2}}{n\sqrt{n^2+x^2}} = \frac{n-\sqrt{n^2+x^2}}{n\sqrt{n^2+x^2}} \frac{n+\sqrt{n^2+x^2}}{n+\sqrt{n^2+x^2}}= \frac{-x^2}{\sqrt{n^2+x^2}\left(n+\sqrt{n^2+x^2}\right)}$. It seems to be a good trick, I will look at it. $\endgroup$ – George Oct 20 '16 at 5:48
  • $\begingroup$ Then $f_n(x) = -x^2 g_n(x)$. By the way, the series $g_n$ is bounded as $|g_n| \leq \frac{1}{2n^3}$, isn't it? $\endgroup$ – George Oct 20 '16 at 18:56
  • $\begingroup$ $g_n$ is bounded because $\sum_{n\geq 1}\frac{1}{2n^2}$ is convergent. $\endgroup$ – Jack D'Aurizio Oct 20 '16 at 19:08
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For fixed $x$,

$$f_n(x)=\frac{1}{n}((1+(\frac{x}{n})^2)^{-\frac{1}{2}})-\frac{1}{n}$$

$$=\frac{1}{n}(1-\frac{1}{2}\frac{x^2}{n^2}(1+\epsilon(n)))-\frac{1}{n}$$

$$=-\frac{x^2}{2n^3}(1+\epsilon(n))$$

thus the series $\sum f_n(x)$ converges.

Now let $a>0$.

for large enough $n$

$\forall x\in [-a,a]$

$|f_n(x)|\leq \frac{a^2}{2n^3}$

thus the series converges uniformly at

$ [-a,a]$ and also at $[a,b]$ but not on $\mathbb R.$

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  • $\begingroup$ You are tacitly assuming that the radius of convergence of the Taylor series of $\frac{1}{\sqrt{1+x^2}}$ at $x=0$ is $+\infty$, that is false. You cannot prove or disprove uniform convergence through that. $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 18:17
  • $\begingroup$ i used equivalence $\endgroup$ – hamam_Abdallah Oct 19 '16 at 18:52
  • $\begingroup$ My objection still holds. You used a Taylor expansion at $0$, that does not converge to the actual value of the function if $x$ is too large in absolute value. Equivalence holds in a neighbourhood of zero, not on the whole real line. $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 18:56
  • $\begingroup$ no, Taylor expansion when $n\to \infty$ $\endgroup$ – hamam_Abdallah Oct 19 '16 at 19:05
  • $\begingroup$ Still the same objection. $\frac{1}{\sqrt{x^2+n^2}}$ is not an entire function, regarded as a function of $x$ or $n$. $\endgroup$ – Jack D'Aurizio Oct 19 '16 at 19:08

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