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Is $$\sum_{n=1}^{\infty}\frac{1}{n(1+(x-n)^2)}$$ uniformly convergent on $(0,+\infty)$? Sadly, Weierstrass' criterion nor Cauchy's condition don't seem to work here and I just ran out of ideas.

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  • $\begingroup$ How about comparing it to $\sum_{n=1}^{\infty}\frac{1}{n^2}$? I think that there must $(\exists n_0)(\forall n>n_0)1+(x-n)^2>n \implies \frac{1}{n(1+(x-n)^2)}<\frac{1}{n^2}$? $\endgroup$ – mirgee May 10 '14 at 9:30
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Yes, indeed uniformly convergent for $x\in\Bbb R$. You need to show that for every $\varepsilon>0$, there exists $n_0=n_0(\varepsilon)$ for which the tail $\sum_{n>n_0(\varepsilon)} 1/n(1+(x-n)^2)$ is less than $\varepsilon$ for all $x$. But $$ \sum_{n>n_0(\varepsilon)} \frac1{n(1+(x-n)^2)} < \frac1{n_0(\varepsilon)} \sum_{n>n_0(\varepsilon)} \frac1{1+(x-n)^2} < \frac1{n_0(\varepsilon)} \sum_{n\in\Bbb Z} \frac1{1+(x-n)^2}, $$ and this last sum (it can be checked) is a continuous function of $x$ that is periodic, hence absolutely bounded, say by $B$. Therefore choosing $n_0(\varepsilon) > B/\varepsilon$ suffices.

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