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Consider the following series of functions: $\sum\limits_{n=1}^{\infty} \frac{nx}{1+n^5x^2}$ on all of $\mathbb{R}$. The first thing we need to do is check for pointwise convergence of the series. To do so, we fix $x$ and see if the series converges.

We obtain: $|f_n(x)| = |\frac{nx}{1+n^5x^2}| \leq \frac{1}{|x|n^4}$ and hence $\sum\limits_{n=1}^{\infty} f_n(x)$ converges pointwise on $\mathbb{R}$ by the comparison test.

Now to check for uniform convergence, we would like to use the Weierstrass test. In class, my professor does the following: $|\frac{nx}{1+n^5x^2}| = \frac{1}{n^{3/2}}|\frac{n^{5/2}x}{1+n^5x^2}| < \frac{1}{2n^{3/2}} = M_n$ and this is true for all $x \in \mathbb{R}$. Since $\sum\limits_{n=1}^{\infty}M_n$ converges, then $\sum\limits_{n=1}^{\infty} f_n(x)$ converges uniformly on $\mathbb{R}$ by the Weierstrass test.

My question is, why does the professor go through all that trouble to find an $M_n$ when it appears that we have a valid one $\frac{1}{|x|n^4}$ when trying to see if the series converges pointwise on $\mathbb{R}$?

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    $\begingroup$ $M_n$ has to depend only on $n$. You could get rid of $|x|$ in $\frac{1}{|x|n^4}$ but it wouldn't yield uniform convergence on the whole real line. $\endgroup$ – Gabriel Romon Apr 20 '15 at 4:23
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Use calculus to find themaiimum of $|f_n|$ on $(-\infty,\infty)$. Since $f_n$ is odd it is enough to stay $(0,\infty)$. $$ f_n'(x)=\frac{n-n^6x^2}{(1+n^5x^2)^2}=0\implies x=n^{-5/2} $$ and $$ |f_n(x)|\le f_n(n^{-5/2})=\frac12\,n^{-3/2}\quad\forall x\in(-\infty,\infty). $$

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