confused by tutorial/example on uniform convergence of $\zeta(x)=\sum 1/n^x$

Question

I am confused by a tutorial/example on uniform convergence of $\zeta(x)=\sum 1/n^x$.

The source is a reputable and widely used website: https://brilliant.org/wiki/uniform-convergence/. An screenshot is replicated below for convenience. Note: the example given is for real, not complex, $x$.

The steps are as follows:

• The Weierstrass M-test says that if $\sum M_n<\infty$ where $|f_n(x)|\leq M_n$, then $\sum f_n(x)$ converges uniformly.
• Fix $a>1$, and note that $|1/n^x|=1/n^x \leq 1/n^a$ where $x\geq a$.
• So the M-test tells us that $\sum 1/n^x$ converges uniformly for $x \geq a$.

Note that I have said $x\geq a$, not $x>a$ as per the tutorial. I think I am correct (?).

This logic can be extended easily from real $x$ to complex $s$. That is, $\sum 1/n^s$ converges uniformly for $\Re(s) \geq a$, where $a>1$.

Question: why does this tutorial, and other sources, not say the uniform convergence is over $x>1$. Why do they separate the conditions into $x \geq a$ and $ a>1$? In my mind the two are equivalent.

The context for this question is that uniform converge of a series where each term is holomorphic means the infinite sum is also holomorphic, and I wan to be able to use this to say the Dirichlet series for $\zeta(s)$ is holomorphic for $\Re(s)>1$

Note: My previous questions have not been well formulated, so I am hoping this one is.

Answer 1

THe reason why they're not equivalent is because of the dependencies of quantifiers. Let $\{f_n\}_{n=1}^{\infty}$ be a sequence of complex-valued functions defined on $(1,\infty)$, and let $f:(1,\infty)\to\Bbb{C}$ be a given function.

• $\{f_n\}_{n=1}^{\infty}$ is uniformly convergent on $(1,\infty)$ to $f$ by definition means the following: for every $\epsilon>0$, there is an $N\in\Bbb{N}$ such that for all $n\in\Bbb{N}$, if $n\geq N$ then $\sup\limits_{x\in (1,\infty)}|f_n(x)-f(x)|<\epsilon$.
• For every $a>1$, $\{f_n\}_{n=1}^{\infty}$ converges uniformly to $f$ on $[a,\infty)$ means by definition that: for every $a>1$, for every $\epsilon>0$, there is an $N\in\Bbb{N}$ such that for all $n\in\Bbb{N}$, if $n\geq N$ then $\sup\limits_{x\in (a,\infty)}|f_n(x)-f(x)|<\epsilon$.

Note that in the first case, $N$ depends only on $\epsilon$, but in the second case, it depends on $a$ and $\epsilon$. Therefore, the first condition implies the second, but the converse need not be true (and it's actually false).

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We don't even have to talk about uniform convergence to highlight the mistake in logic you're making. Here are much simpler statements, which hopefully highlight the error in your logic:

False Statement 1.

Every natural number is finite, therefore $\Bbb{N}$ is a bounded set.

I hope no explanations are necessary as to why it is false. Here's another (essentially same logical error, just slightly more sophisticated):

False Statement 2.

Consider the function $f:(0,\infty)\to\Bbb{R}$ defined as $f(x)=\frac{1}{x}$. For every $\alpha>0$, $f$ is bounded on $[\alpha,\infty)$ (it is bounded by $\frac{1}{\alpha}$ since it is a decreasing function). Therefore, $f$ is bounded on $(0,\infty)$.

Clearly this is an incorrect conclusion. Just sketch the graph of $f$ to see it is unbounded on $(0,\infty)$. Again, this is a matter of dependencies of quantifiers in the definition.

• To say $f$ is bounded on $(0,\infty)$ means that there is an $M>0$ such that for all $x\in (0,\infty)$, $|f(x)|\leq M$.
• To say for each $\alpha>0$, $f$ is bounded on $[\alpha,\infty)$ means that for every $\alpha>0$ there is an $M>0$ such that for all $x\in [\alpha,\infty)$, we have $|f(x)|\leq M$.

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The issue is that things like "boundedness" or "uniform convergence" are "global" properties, depending on the behavior of functions (or sequences of functions) on the whole domain of of the functions. Contrast this with things like "continuity" or "differentiability", which are "local properties", in the sense that if a function is continuous (resp differentiable, resp $C^{\infty}$) on $(a,\infty)$ for all $a>1$, then the function is continuous (resp differentiable, resp $C^{\infty}$) on $(1,\infty)$.