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
 A: There are two aspects to your question:


*

*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".

*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.
