# The Series of functions in Weierstrass M test

I have some questions regarding the sequences of functions in the Weierstrass M-test:

Weierstrass M-test:. Suppose that $\{f_n\}$ is a sequence of real- or complex-valued functions defined on a set $A$ .....

My question is specifically regarding the $f_n(x)$ in the definition. From all the example that I have seen so far, the sequence of function is usually related by the index. For example, one of the example I seen is:

$g_n(x) = \frac{x^n}{n}$

So it means $g_1(x) = \frac{x^1}{1}, g_2(x) = \frac{x^2}{2}, g_3(x) = \frac{x^3}{3}$

I am just wondering does the sequence of functions i.e. $f_1(x), f_2(x), f_3(x), ...$

have to be related like the above example? Or it could be like this:

$f_1(x)= \sin(x), f_2(x)=\cos(x), f_3(x)= x^2+x + 11, f_4(x)=\frac{3^x}{5}$, ...

Because when I was looking at the proof of the theorem, it seems it is using the partial sum of each of the above functions i.e. looking at $S_n(x)=f_1(x)+f_2(x)+f_3(x)+...+f_n(x)$

Does the $f_1(x), f_2(x),... f_n(x)$ have to be related? or they could be totally arbitary different function defined on the same domain?

Thank you

• They can be arbitrary, as the theorem states. – user42761 Dec 23 '16 at 11:56
• if you look at the proof, it never assumes form of $f_n$ to be related to $n$ in any way – user160738 Dec 23 '16 at 12:00

There are two aspects to your question:

1. The functions $f_{n}$ in the $M$-test are assumed to satisfy $$\sup_{x \in A} |f_{n}(x)| = M_{n},\qquad \sum_{n} M_{n} < \infty. \tag{*}$$ That is a relationship between $f_{n}$ and the index $n$, albeit one that's particularly weak. In this sense, it's inaccurate to say the $f_{n}$ are "totally arbitrary".

2. You can apply the Weierstrass $M$-test to a sequence $(f_{n})$ that starts off with the functions you mention. In order for a function sequence to be defined, however, $f_{n}$ has to be specified for all $n$. In practice, that's usually achieved by having $f_{n}(x)$ be given by a closed formula involving $n$. In this sense, you're unlikely to encounter $$\sin x + \cos x + (x^{2} + x + 11) + \tfrac{3^{x}}{5} + \cdots$$ in practice, and very likely to see, e.g., $$\sum_{n=0}^{\infty} \frac{x^{n}}{n!},\qquad \sum_{n=1}^{\infty} e^{-nx},\qquad \sum_{n=1}^{\infty} \frac{\sin(nx)}{n^{2}}.$$

In case a bit of theory helps: Convergence of an infinite series is entirely determined by asymptotic behavior of the tail, not by any finite number of terms. To give a precise formulation, if $(a_{n})_{n=1}^{\infty}$ is a sequence (of complex numbers, say), then following are equivalent:

• For some $N \geq 1$, $\displaystyle\sum_{n=N}^{\infty} a_{n}$ converges.

• For every $N \geq 1$, $\displaystyle\sum_{n=N}^{\infty} a_{n}$ converges.

Consequently, condition (*) in the $M$-test is not a condition on any finite collection of the $f_{n}$, but a condition on the asymptotic behavior of the suprema. You can "fold in" or remove an arbitrary finite set of terms without changing whether or not the series converges.

Or, if you like, in the Weierstrass $M$-terst, an arbitrarily long initial finite sequence of terms is completely arbitrary, even though the asymptotic behavior of the tail is not.

• Thank you Dr. Hwang for your explanations. I think it is quite clear. But I just have one more question, since you mentioned that the M-test is more concerned about the tail behavior, does it mean we would never have something like this: $\frac{x}{x^n}$ at the end of the series of function? but it is possible to have those terms say for the first 855 terms in the series? but some other sequences of functions that would converge after the 855th functions in the sequences? Say we can have the sequence of functions being: – john_w Dec 29 '16 at 11:03
• $g_1(x) = \frac{x}{x^1}$, $g_2(x) = \frac{x}{x^2}$, ..., $g_{855}(x) = \frac{x}{x^{855}}$, $g_{856}(x) = \frac{x^1}{1!}$, $g_{857}(x) = \frac{x^2}{2!}$, $g_{858}(x) = \frac{x^3}{3!}$, .... as you mentioned above? thank you – john_w Dec 29 '16 at 11:10
• The "tail" of a series refers generically to the terms remaining after finitely many terms have been dropped, so "finitely many terms in the Weierstrass $M$-test are arbitrary" means "folding in finitely many terms has no effect on uniform convergence". On the other hand, for the $M$-test to make sense, each summand $f_{n}$ must be bounded. If $A$ is the real line (or the set of non-zero real numbers), a function such as $g_{2}(x) = x/x^{2}$ can't arise, since that function isn't bounded on $A$. [...] – Andrew D. Hwang Dec 29 '16 at 13:23
• For the functions in your second comment, you do have that the sequence of partials sums starting with the $856$th partial sum (i.e., including all the singular terms) converges uniformly to the sum of the entire series. The $M$-test itself doesn't apply to the entire series, but does show the convergence$$f_{n}(x) = \sum_{k=856}^{856+n} g_{k}(x) \to \sum_{k=856}^{\infty} g_{k}(x)$$is uniform, so that$$\sum_{k=0}^{856+n} g_{k}(x) \to \sum_{k=0}^{\infty} g_{k}(x)$$is also uniform ("in $n$"), even though the terms of the sequence and the limit may be unbounded on $A$. – Andrew D. Hwang Dec 29 '16 at 13:30
• Thank you Dr. Hwang. I think I understand much much more and more clear in the understanding of the M-test and the general convergence of series (i.e. the focus of asymptotic behaviour). Happy New Year 2017! – john_w Dec 31 '16 at 3:58