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From Understanding Analysis by Abbott:

Exercise 6.4.2 (c): If $\sum_{n=1}^{\infty} f_n$ converges uniformly on $A$, then there exist constants $M_n$ such that $|f_n(x)| \leq M_n$ for all $x \in A$ and $\sum_{n=1}^{\infty} M_n$ converges.

My understanding is that to disprove this by counterexample we need to show that for all $M_n$ such that $|f_n(x)| \leq M_n$ for all $x \in A$ we have that $\sum_{n=1}^{\infty} M_n$ diverges. Obviously we also need that $\sum_{n=1}^{\infty} f_n$ converges uniformly on $A$.

However, most responses online which discuss the converse of the Weirstrass M-test, which I believe this is, (e.g. here Q3b and here) give a particular value of $M_n$ for which $\sum_{n=1}^{\infty} M_n$ diverges.

Is this a difference in the statement of the theorem? Chiefly whether we assume we are given the $M_n$ or not?

Arnold Davidson
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    In the answers in the second link they use the fact that if $\sum M_n$ converges for some dominating sequence $(M_n)$ then the same must be true when $M_n=\sup_x |f_n(x)|$ . Isn't this obvious? – geetha290krm May 08 '23 at 11:45

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The example you cite, is not "a particular" value of $M_{n}$ but a very "special" one.

Weirestrass $M$ test can be equivalently formulated as , if $\displaystyle\sum_{n\geq 1}\sup_{x\in A}|f_{n}(x)|<\infty$ , then $\sum_{n\geq 1}f_{n}(x)$ converges uniformly.

So, in particular if $f_{n}$'s are constant functions $\frac{(-1)^{n}}{n}$ , then $\sup|f_{n}(x)|=\frac{1}{n}$ , the sum of which diverges. But $\sum_{k=1}^{n}f_{k}$ converges(and hence converges uniformly as $f_{n}'s$ are constants) . The main point being that any other constans $M_{n}$ such that $|f_{n}|\leq M_{n}$ would necessarily have to satisfy that $M_{n}\geq\frac{1}{n}$ and hence $\sum_{n}M_{n}$ would also necessarily diverge for all such $M_{n}$'s .

See @Jyrki's example in the comments below for a more interesting example rather than just constant $f_{n}$'s.

Mr.Gandalf Sauron
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    This is my interpretation of the question as well (+1). A slightly more interesting example with the same sequence of constants $M_n$ could be the series $$x-\frac x2+\frac{x^3}3-\cdots=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}n.$$ It converges uniformly in the interval $x\in[0,1]$ to $\ln(1+x)$. All because the alternating series cut-off-error-estimate tends to zero uniformly. – Jyrki Lahtonen May 08 '23 at 11:49
  • @JyrkiLahtonen Ofcourse ! . I just mentioned the confusion of the op with the example he cited . I have edited my answer and referenced your comment. – Mr.Gandalf Sauron May 08 '23 at 11:53
  • Thank you @Mr.GandalfSauron! So, assuming I didn't know about the equivalent formulation (which I didn't, it's not in the textbook) - is this still the most fruitful way (I know this isn't well-defined, but hopefully you get my intent) to approach this problem? – Arnold Davidson May 08 '23 at 12:04
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    @ArnoldDavidson . As I said, any bound $M_{n}$ must satisfy that $\sup|f_{n}|\leq M_{n}$ and also $|f_{n}(x)|\leq \sup_{x\in A} |f_{n}(x)|$ for all $x$ by definition of supremum. So if $\sum_{x}\sup|f_{n}|$ converges then you already have found for $M_{n}=\sup_{x}|f_{n}(x)|$ which converges and you can apply $M$ test. And if for some $M_{n}$'s it hold. Then $\sup_{x}|f_{n}(x)|\leq M_{n}$ implies that $\sum_{n}\sup_{x}|f_{n}(x)|$ must also converge. – Mr.Gandalf Sauron May 08 '23 at 12:09
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    @ArnoldDavidson Also , if say $f_{n}$'s were differentiable and positive functions . Then too , you can do the first order derivative test to find the maximum of $f_{n}$ and then show that the sum of the maximums converge. That is a viable strategy to keep in mind while applying M test. – Mr.Gandalf Sauron May 08 '23 at 12:11