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I'm interested in knowing some values of the zeta function at infinite inputs. If we accept that $$\zeta(\infty) = 1$$ (as shown here in WolframAlpha) then, is it possible to evaluate the zeta function at a divergent series? To be specific, let's take the harmonic series.

Question. What is the value of $$\zeta \left( \sum_{n=1}^{\infty}\frac{1}{n} \right)$$ and is it equal to $\zeta(\infty)$?

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    $\begingroup$ Neither $\zeta\left(\sum\frac{1}{n}\right)$ nor $\zeta(\infty)$ are actually well-defined. $\endgroup$ – Wojowu Nov 5 '16 at 19:32
  • $\begingroup$ @Wojowu True, but improper integrals have no actual value, only a limit, but often , it is written $\int_0^{\infty}\cdots = \cdots$. We could define $\zeta(\infty)=1$ as well in a similar manner. I do not know however, if such a definition is common. $\endgroup$ – Peter Nov 5 '16 at 19:38
  • $\begingroup$ I think if it is not defined at infty wolfram alpha able to show that going to show "Undefined" $\endgroup$ – zeraoulia rafik Nov 5 '16 at 19:39
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    $\begingroup$ @Peter Improper integrals do have a value, which is defined to be the limit you mention, but I see your point. $\endgroup$ – Wojowu Nov 5 '16 at 19:39
  • $\begingroup$ I added the tag "soft-question" because I am not sure whether it is clear whether the expressions here are well defined or not. $\endgroup$ – Peter Nov 5 '16 at 19:47
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There is nothing fancy going on here. If $$H_n = \sum_{k=1}^n \frac{1}{k}$$ then $\displaystyle \lim_{n \to \infty} H_n = \infty$ and since $\zeta(s) \to 1$ as $s$ goes to infinity along the real line, we conclude $$\lim_{n \to \infty} \zeta(H_n) = 1.$$ I believe this is the most obvious interpretation of your symbols.


Given the consternation in the comments, I feel I need to mention the following:

  1. The series $\sum_{k=1}^\infty \frac{1}{k}$ is a series of non-negative real numbers. These are the simplest kinds of series: the associated sequence of partial sums is either bounded or unbounded, and so the series converges to a finite real number or the series diverges; this type of divergence can be interpreted as convergence to $+\infty$ (distinguished from $-\infty$) in the topological space $\overline{\mathbb{R}}$ (the affinely extended real number line).
  2. Whereas the real line has two natural points at infinity, the complex plane has only one, and it is obtained via the Alexandroff a.k.a. one-point compactification of $\mathbb{C}$ to $\overline{\mathbb{C}}$. The resulting space is called the Riemann sphere or the extended plane and it is homeomorphic to the real sphere $S^2$ via stereographic projection and also to complex projective space, $\mathbb{C}P^1$ (notations vary). (Note that the one-point compactification of $\mathbb{R}$ is not the space $\overline{\mathbb{R}}$ in item 1.) On the Riemann sphere, convergence to infinity means eventual departure from any given compact subset of $\mathbb{C}$; thus "$\displaystyle \lim_{z \to \infty} f(z) = L$" means "for all $\epsilon > 0$ there exists $M > 0$ such that $|f(z) - L| < \epsilon$ for all $|z| > M$" for finite $L$; and "$\displaystyle\lim_{z \to \infty} f(z) = \infty$" means "for all $N > 0$ there exists $M > 0$ such that $|f(z)| > N$ for all $|z| > M$".
  3. The obvious interpretation of a real number $x$ as a complex number $x + i0$ is a topological embedding $\iota:\mathbb{R} \to \mathbb{C}$, and $\iota$ extends to the compactifications, yielding a map $\iota^*:\overline{\mathbb{R}} \to \overline{\mathbb{C}}$ which is no longer injective: $\iota^*(+\infty) = \iota^*(-\infty) = \infty$.
  4. When faced with an expression involving a function evaluated at a point outside of that function's domain, it is standard in analysis to interpret the expression as meaning the value of some continuous or analytic extension of the function at that point. This is reasonable whenever said extension is unique w.r.t. the property of being continuous or analytic, and often the issue is not even explicitly addressed. Thus, if $f(z) = \frac{\sin z}{z}$ then $f(0)$ means $1$ because there's nothing else it could reasonably be; $\displaystyle \lim_{z \to 0} \tfrac{\sin z}{z} = 1$ and extending $f$ in this manner results in a continuous, even an analytic, function. Another common example is $g(z) = \frac{e^z - 1}{z}$ with $g(0) = 1$. Yet another is $h(x) = x^{-2}$ with $h(0) = +\infty$ (in $\overline{\mathbb{R}}$). Likewise, $\zeta(0)$ means $-\frac{1}{2}$, even though $\zeta$'s typical domain of definition begins with the half-plane $\operatorname{Re} s > 1$ (coming from the Dirichlet series $\sum_{n \ge 1} n^{-s}$) before being extended to $\mathbb{C}\setminus\{1\}$ via the analytic continuation of $\zeta$.
  5. A suggested edit to my answer added the statement that $\zeta(1) = \infty$, but this is false. It is true that $\zeta$ has a pole at the point $1$, but this only means $\displaystyle \lim_{s\to 1} |\zeta(s)| = \infty$. In fact, the limit without the absolute value bars does not exist: $$\lim_{s \to 1^+} \zeta(s) = \infty$$ whereas $$\lim_{s \to 1^-} \zeta(s) = -\infty$$ as $s$ goes through real values; see this graph.
  6. By items 3 and 5, we interpret $\zeta(\infty)$ to mean $\displaystyle \lim_{s \to \infty} \zeta(s)$ as a limit on the Riemann sphere. Alas, this limit does not exist. As I already pointed out, $\zeta(s) \to 1$ as $s \to \infty$ through reals, but $$\displaystyle \lim_{\substack{s\to -\infty \\ s\in\mathbb{R}}} \zeta(s)$$ does not exist because $\zeta$ oscillates wildly for large negative $s$, thanks to the sine and Gamma terms in the functional equation. Limits along other directions of escape may or may not exist (e.g. people talk about $i\infty$, and $\displaystyle \lim_{t \to \infty} \zeta(1 + it) = 0$, if I'm not mistaken), but the fact that already these few "sub"-limits don't agree means there is no continuous extension of $\zeta$ to $\infty$.
  7. By the way, there do exist complex functions with continuous extensions at $\infty$; $f(z) = \frac{1}{z}$ is one, as are other Mobius transformations. But if $f$ is entire and $\displaystyle \lim_{z \to \infty} f(z)$ exists, then $f$ is constant.

Now we are in a position for a satisfactory resolution of your problem. First, we interpret $$\sum_{n=1}^\infty \frac{1}{n} = +\infty \in \overline {\mathbb{R}}.$$ In my answer above, I implicitly considered $\zeta$ as a function of a real variable only, because this does possess a continuous extension to $(1, \infty) \cup \{+\infty\}$. We can evaluate $\zeta(+\infty)$ by taking the limit of $\zeta$ evaluated at the partial sums $H_n$, as I did above, and under this interpretation, $$\zeta \left( \sum_{n=1}^\infty \frac{1}{n} \right) = 1.$$

However, if we are to consider $\zeta$ truly for what it is—a function of a complex variable—then we should interpret the input accordingly: either as a complex number, $$\iota^*\left(\sum_{n=1}^\infty \frac{1}{n}\right) = \iota^*(+\infty) = \infty \in \overline{\mathbb{C}},$$ or as the sum of a series of complex numbers, $$\sum_{n=1}^\infty \frac{1}{n} = \infty \in \overline{\mathbb{C}}.$$ In either case, $$\zeta \left( \sum_{n=1}^\infty \frac{1}{n} \right) \text{ does not exist.}$$

In conclusion, whether or not we have the "identity" $$\zeta \left(\sum_{n=1}^\infty \frac{1}{n} \right) = \zeta(\infty)$$ essentially depends on your interpretation of $\infty$.


And finally, the answer to your more general question "is it possible to evaluate $\zeta$ at a divergent series" is "it depends on the interpretation of the series".

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    $\begingroup$ @Peter I think you're right, I misviewed that...anyway, that limit is wrong, I think. $\endgroup$ – DonAntonio Nov 5 '16 at 20:07
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    $\begingroup$ @user51189 Perhaps that's true, yet to take WA as base for a justification is, I think, a rather weak argument. Almost anyone who's been a member of this site had the opporunity to see WA has big mistakes sometimes. Who knows what interpetration is that program taking $\endgroup$ – DonAntonio Nov 5 '16 at 20:08
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    $\begingroup$ I admit I did not realize that one particular term of my sequence is undefined (namely, the $n = 1$ term, which is $\zeta(1)$). However, it is common practise in analysis to disregard finitely many initial terms of sequences when considering their limits. Moreover, I did not claim that $\lim_{s \to \infty} \zeta(s)$ exists as a complex limit (so $|s| \to \infty$ in any direction); this is a much stronger statement than mine (and it is false). $\endgroup$ – Unit Nov 5 '16 at 20:27
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    $\begingroup$ @Peter It is the same: if as written in the above answer the intention is to take the limit of the zeta function when $\;s\to\infty,\,s\in\Bbb R\;$ , then it must be proved the limit is infinity. Observe the OP wrote $$\;\zeta\left(\sum_{n=1}^\infty\frac1n\right)\;$$ which, imo, seems to be a mathematical nonsense as the argument of the zeta function doesn't even exist as defined in real series...but even if the intention was to take the limit, as this answer means, things must be proved: $$\text{is it true that}\;\;\;\lim_{x\to\infty}\frac1{n^x}=\infty.....??$$ That is my point. $\endgroup$ – DonAntonio Nov 5 '16 at 20:28
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    $\begingroup$ I'll end this long and rather interesting thread by pointing out that iff the limit exists then it seems (to me, at least) to be $\;1\;$ because of the values of $\;\zeta(2n)\;$ . So far I haven't found anything else on this in all my stuff, perhaps later I shall make a search in the web or whatever. Thanks. $\endgroup$ – DonAntonio Nov 5 '16 at 20:32

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