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Problem: Prove that the following series converges uniformly on the given interval using Weierstrass M-test. $$\sum\frac{nx}{1+n^5x^2} $$ on $|x| < +\infty$. My attempt has been to use $M_{n} = \frac{nx}{n^5x^2} = \frac{1}{n^4 x} $, and show that this converges using the comparison test, but then I get interval of convergence $|x| > 1$. Do I have the wrong idea about how to use the M-test?

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  • $\begingroup$ I found an easier way to prove the uniform convergence $\endgroup$ – Raffaele Apr 1 '14 at 10:17
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You can not take take $M_{n} = \frac{nx}{n^5x^2} = \frac{1}{n^4 x}$ since when $x=0$, $M_n$ is not bounded. If you calculate the supremum value of $\frac{nx}{1+n^5x^2}$ then you will see that it takes its supremum at $x=n^{-5/2}$. Then you get $M_{n}=\frac{n^{-3/2}}{2}$ and $\sum M_{n}=\sum\frac{n^{-3/2}}{2} $ is convergent , so by Weierstrass M-test the series converges uniformly.

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  • $\begingroup$ Why didn't you get $\sum{M_n}=\sum{n^{-5/2}}$? $\endgroup$ – Raffaele Apr 1 '14 at 10:19
  • $\begingroup$ at $x=n^{-5/2}$, $\frac{nx}{1+n^5x^2}$ takes its supremum and its supremum is $$M_n=\frac{n.n^{-5/2}}{1+n^5\left(n^{-5/2}\right)^2}=\frac{n^{-3/2}}{2}$$ $\endgroup$ – guest Apr 1 '14 at 10:27
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The misunderstanding in your way of using the Weierstrass criterium is that $M_n$ must not depend on $x$. If the proof below uses concept that you have not studied yet, keep in mind that the Weierstrass M-criterium needs that all functions have to be less than a constant $M_n$ on all $\mathbb{R}$ and the series of these $M_n$ must be convergent.

The maximum on all $\mathbb{R}$ of each function $f_n(x)$ is at the $x$ where $f'(x)=0$ $${{f'}_n}\left( x \right) = \frac{{n\left( {1 - {n^5}{x^2}} \right)}}{{{{\left( {1 + {n^5}{x^2}} \right)}^2}}}$$ $${{f'}_n}\left( x \right) = 0 \to 1 - {n^5}{x^2} = 0 \to x = \pm \frac{1}{{{n^{\frac{5}{2}}}}}$$ $${f_n}^{\prime \prime }\left( x \right) = \frac{{2{n^6}x\left( {{n^5}{x^2} - 3} \right)}}{{{{\left( {1 + {n^5}{x^2}} \right)}^2}}};\quad {f_n}^{\prime \prime }\left( {\frac{1}{{{n^{\frac{5}{2}}}}}} \right) = - \frac{{{n^{\frac{7}{2}}}}}{2} < 0$$ Thus for any $n$ we have on all $\mathbb{R}$ that $f_n(x)\le \frac{1}{n^{\frac{5}{2}}}$ and as $\sum {\frac{1}{n^{\frac{5}{2}}}} $ converges for Dirichlet criterium, the given series converges uniformly on $\mathbb{R}$.$\square$

You can use a generalized version of Weirstrass theorem, which holds in Banach spaces a function series $\sum {{f_n}\left( x \right)} $ is uniformly convergent if for any index $n$ we have $\left\| {{\kern 1pt} {f_n}{\kern 1pt} } \right\| \leqslant {M_n}$ and $\sum {{M_n}}$ converges.

If we define in $\mathbb{R}$ the norm of a function $f(x)$ as $$\left\| {{\kern 1pt} f\left( x \right){\kern 1pt} } \right\| = {\left[ {\int\limits_\mathbb{R} {{f^2}} \left( x \right){\text{ d}}x} \right]^{\frac{1}{2}}}$$ we have for any function $f_n(x)$ of the given series $$\left\| {{\kern 1pt} {f_n}\left( x \right){\kern 1pt} } \right\| = {\left[ {{{\int\limits_\mathbb{R} {\left( {\frac{{nx}}{{1 + {n^5}{x^2}}}} \right)} }^2}{\text{ d}}x} \right]^{\frac{1}{2}}}$$ A primitive is $${\int {\left( {\frac{{nx}}{{1 + {n^5}{x^2}}}} \right)} ^2}{\text{ d}}x = \frac{{\arctan \left( {{n^{\frac{5}{2}}}x} \right)}}{{2{n^{\frac{{11}}{2}}}}} - \frac{x}{{2{n^3}\left( {1 + {n^5}{x^2}} \right)}} + C$$ therefore $${\int\limits_\mathbb{R} {\left( {\frac{{nx}}{{1 + {n^5}{x^2}}}} \right)} ^2}{\text{ d}}x = \mathop {\lim }\limits_{h \to \infty } \left[ {\frac{{\arctan \left( {{n^{\frac{5}{2}}}x} \right)}}{{2{n^{\frac{{11}}{2}}}}} - \frac{x}{{2{n^3}\left( {1 + {n^5}{x^2}} \right)}}} \right]_{ - h}^h = \frac{\pi }{{2{n^{\frac{{11}}{2}}}}}$$ and so $$\left\| {{\kern 1pt} {f_n}\left( x \right){\kern 1pt} } \right\| = {\left( {\frac{\pi }{{2{n^{\frac{{11}}{2}}}}}} \right)^{\frac{1}{2}}} = \sqrt {\frac{\pi }{2}} \frac{1}{{{n^{\frac{{11}}{4}}}}}$$ As the series $$\sqrt {\frac{\pi }{2}} \sum\limits_{n = 1}^\infty {\frac{1}{{{n^{\frac{{11}}{4}}}}}} $$ is convergent for Dirichlet criterium, the series of function is uniformly convergent over $\mathbb{R}$.

Hope this can help you...

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