Is this assumption correct in proof of Weierstrass M-Test in Spivak's Calculus?

Question

The following is a theorem in Spivak's Calculus, Chapter 24 entitled "Uniform Convergence and Power Series"

Theorem (Weierstrass M-Test)

Let $\{f_n\}$ be a sequence of functions defined on $A$ .

Suppose $\{M_n\}$ is a sequence of numbers such that $$|f_n(x)|\leq M_n\ \ \text{ for all } x\in A$$

Suppose that $\sum\limits_{n=1}^\infty M_n$ converges.

Then, for each $x\in A$ the series $\sum\limits_{n=1}^\infty f_n(x)$ converges (in fact, it converges absolutely) and $\sum\limits_{n=1}^\infty f_n$ converges uniformly on $A$ to the function

$$f(x)=\sum\limits_{n=1}^\infty f_n(x)$$

My question is about the assumption that $\sum\limits_{n=1}^\infty M_n$ converges.

As written, this series can converge to any value.

The proof of this theorem goes as follows

Let $x\in A$ .

Consider the sequence $\{|f_n(x)|\}$ , each term of which is smaller than the corresponding term in the sequence $\{M_n\}$ .

Since $\sum\limits_{n=1}^\infty M_n$ converges, ie since $\{M_n\}$ is summable, then by the comparison test $\{|f_n(x)|\}$ is summable, ie $\sum\limits_{n=1}^\infty |f_n(x)|$ converges.

For all $x\in A$ we have

$$\left |f(x)-\sum\limits_{n=1}^N f_n(x)\right | = \left | \sum\limits_{n=N+1}^\infty f_n(x) \right |$$ $$\leq \sum\limits_{n=N+1}^\infty |f_n(x)|$$ $$\leq \sum\limits_{n=N+1}^\infty M_n$$

Since $\sum\limits_{n=1}^\infty M_n$ converges then the number $\sum\limits_{n=N+1}^\infty M_n$ can be made as small as desired by choosing $N$ sufficiently large.

In this last sentence, doesn't $\sum\limits_{n=1}^\infty M_n$ have to converge to 0?

Answer 1

$\sum_{n = 1}^\infty M_n$ need not converge to 0, and in general note that it will not, since the condition $|f_n(x)| \leq M_n$ for all $x \in A$ for each $n$ implies each $M_n$ is non-negative, so you are taking a series of non-negative terms, and thus it will be positive if even one term is positive.

You can get an explicit counterexample in any common example of the application of the M-test, one common one is to prove a series like $\sum_{n = 1}^\infty \frac{1}{x^2 + n^2}$ converges, where we take $\{M_n\}_{n = 1}^\infty = \{\frac{1}{n^2}\}_{n = 1}^\infty$, and certainly $\sum_{n = 1}^\infty M_n \neq 0$.

Answer 2

In this last sentence, doesn't $\sum\limits_{n=1}^\infty M_n$ have to converge to 0?

But the series in that last sentence, that not start at $1$. It starts at $N+1$. A series is convergent if and only if its tails can be small (this is simply a restatement of the fact that the sequence of partial sums has to be Cauchy). The fact that $\sum_nM_n<\infty$ (and this is a series of non-negative terms) implies by definition that given $\varepsilon>0$ you can find $N$ large enough such that $\sum_{n=N+1}^\infty M_n<\varepsilon$.

The above tells you absolutely nothing about the value of the series $\sum_nM_n$. In fact, and this is crucial to understanding convergence, whatever happens in the first however many terms you wants, is completely irrelevant to the existence of the limit. Only the behaviour "at infinity" matters.

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Edit: more details.

By definition, that $\sum_nM_n<\infty$ means that the sequence $\{S_k\}$ of partial sums converges. A sequence of real numbers converges if and only if it is Cauchy. This means that given $\def\e{\varepsilon}\e>0$, there exists an index $n_0$ such for all $m>n>n_0$, $|S_m-S_n|<\e$. This we can rewrite as $$ \e>|S_m-S_n|=\Big|\sum_{k=1}^mM_k-\sum_{k=1}^nM_k|=\Big|\sum_{k=n+1}^nM_k\Big|=\sum_{k=n+1}^mM_k. $$ The last equality is justified because the terms are non-negative. As this holds for all $m$ and a monotone bounded sequence is convergent, this is equivalent with $$ \sum_{k=n+1}^\infty M_n<\infty. $$