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Show that the series $$\sum {x\over1+n^2x}$$ is uniformly convergent in $[\delta,1]$ for any $\delta>0$ but not uniformly convergent in $[0,1]$

I am unable to find a suitable $M_n$ for use in the Weirestrass M-test. Need some guidance please.

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  • $\begingroup$ As an aside, $~\displaystyle\sum_{n=-\infty}^\infty\frac1{1+n^2x^2}=\frac\pi x\coth\frac\pi x$. This can be shown by differentiating the natural logarithm of Euler's infinite product for the sine function. $\endgroup$ – Lucian Jun 3 '14 at 19:13
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Are you sure it is not uniformly convergent on all of $[0,1]$? For any $x \in [0,1]$ we have that $f_n(x) = \frac{x}{1+xn^2} \leq \frac{1}{n^2}$. Indeed, for $x \in (0,1]$ we have $f_n(x) = \frac{1}{(1/x) + n^2} \leq \frac{1}{n^2}$ and since $f_n(0)=0$, it is true for all $x \in [0,1]$. Letting $M_n = \frac{1}{n^2}$ would then suggest (via the $M$-test) that $\sum_{n=1}^{\infty} f_n$ is uniformly convergent on the domain $[0,1]$.

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$M_n = \frac{1}{n^2}$ and it is uniformly convergent on $[0,1]$.

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Try $M_{n}=\frac{1}{1+\delta n^2}$.

We have $x \in [\delta,1]$, so:

$\frac{x}{1+n^2x} \leq \frac{1}{1+n^2x}$

It's because $x \leq 1$. Next:

$\frac{1}{1+n^2x} \leq \frac{1}{1+n^2\delta}$ because $\delta \leq x$.

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  • $\begingroup$ how do we get this $M_n$ $\endgroup$ – Aman Mittal Jun 3 '14 at 18:33

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