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Given $g_n(x)= \frac{nx}{1+n^2 x}$, $g_n : [0,1] \to \mathbb R, n \in \mathbb N, n\geq 1$, I am trying to show that $g_n$ converges uniformly. I have shown that it converges pointwise to $g:[0,1]\to\mathbb R$, with $g(x)=0$. Now I wish to show that it converges uniformly.

I know how to do this using the epsilon definition, however I tried applying the M test, which states that if $$\sum_{n=1}^{\infty} ||g_n||_\infty < \infty$$ then the sequence converges uniformly. However, for this sequence, this yields me with $$\sum_{n=1}^\infty ||\frac{nx}{1+n^2x}||_\infty =\sum_{n=1}^\infty \sup_{x\in [0,1]} \left|\frac{nx}{1+n^2x}\right|=\sum_{n=1}^\infty \left|\frac n {1+n^2} \right|.$$ To my knowledge, this is divergent.

On the contrary, we were told that $$\lim_{n\to \infty} ||f_n - f||_\infty = 0 \iff f\text{ is uniformly convergent.}$$ Using this definition here, however, yields $$\lim_{n\to\infty}||\frac{nx}{1+n^2x}-0||_\infty = \lim_{n\to\infty}\frac{n}{1+n^2}=0.$$

I have tried searching yet am unable to clarify where my mistake here lies. Is the M test not an equivalence? Or am I merely mistaken?

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    $\begingroup$ Are you trying to show $(g_n)_{n \in \Bbb{N}}$ converges uniformly, or $\sum_{n=1}^\infty g_n$ converges uniformly? $\endgroup$ – Clement Yung May 4 at 16:34
  • $\begingroup$ @ClementYung I'm trying to show that $(g_n)_{n\in\mathbb N}$ converges uniformly - I mistakenly thought that the M test meant that the sequence will converge. However, I was taught that if the M test holds, ie the summ is <inf, then the sum $\sum_{n=1}^{\infty} f_n$ converges absolutely and uniformly to a function $f:D\to\mathbb R$. $\endgroup$ – YamahaJacoby May 4 at 16:39
  • $\begingroup$ Thank you for your response. Also noting the answer, can I ask for an explanation of what this test actually does then? It was my understanding that this means that the sum of $f_n$ converges to $f$. Is this not the case? @ClementYung $\endgroup$ – YamahaJacoby May 4 at 17:19
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    $\begingroup$ Weierstrass M-test asserts that for each function $g_n$, if $\|g_n\|_\infty < M_n$ for some constant $M_n > 0$, and $\sum_{n=1}^\infty M_n < \infty$, then $\sum_{n=1}^\infty g_n$ converges uniformly to some function $g$. It seems like you've mixed up the convergence of $g_n$ and $\sum_{n=1}^\infty g_n$ here. $\endgroup$ – Clement Yung May 4 at 17:22
  • $\begingroup$ Thank you again. So just for clarity - $\sum_{n=1}^{\infty} g_n$ converges uniformly to some function $g$. But this function, $g$ is not the same as the function that the functionsequence $g_n$ converges to uniformly? (assuming $g_n$ conv. uniformly) $\endgroup$ – YamahaJacoby May 4 at 17:26
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If I understand correctly you are asking if the following implication holds: $$(g_n) \ \text{converges uniformly} \implies\Sigma||g_n||_\infty<\infty $$

To see that this is false take $\forall x\ \ :g_n(x)=\frac{1}{n}$.

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