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The expression

$$ \sum_{n=1}^\infty n = -\frac{1}{12} $$

has been met with considerable controversy. Previous questions (such as this and this) searching for a way of rationalizing the formal convergence of this series in some metric space.

My question is more general. Given a function expressed as a series, for example $\zeta(s) := \sum_{i=1}^\infty n^{-s}$, which converges over some domain $D$, construct its analytic continuation $\bar{\zeta}$ to $\mathbb{C}\setminus S$ where $S$ is a set of singularities. Is there a natural topology on $\mathbb{C}$ in which $\sum_{n=1}^\infty n^{-s}$ converges to $\bar{\zeta}(s)$ for all $s\in \mathbb C \setminus S$?

More specifically, is there a natural topology on $\mathbb{C}$ in which $\sum_{n=1}^\infty n = -\frac{1}{12}$? I am very aware that one can construct a metric in which $\sum_{n=1}^\infty n$ converges to $-1/12$, as is done here. But this is not interesting: one can define a metric so that $\sum_{n=1}^\infty n$ converges to anything.

Given that the analytic continuation of a series is unique, it seems intuitively plausible at least that it is at least in principle possible to define a topology in which the series converges everywhere to its analytic continuation. Can this be done?

eepperly16
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  • What do you mean by "natural"? – Gerry Myerson Nov 16 '17 at 22:01
  • @Gerry Myerson I am using "natural" quite imprecisely because I'm sure what formal requirements I would want such a topology to have. Ideally I'd want something analogous to the weak topology in functional analysis that could simultaneously give convergence to every series to its analytic continuation, while preserving standard Cauchy convergence. I'm not sure whether this is possible so I'd be interested in any topology in which, for example, $\sum n^{-s} = \bar{\zeta}(s)$ whose construction doesn't feel horribly contrived – eepperly16 Nov 16 '17 at 22:11
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    There is no such natural topology. Nevertheless $\eta(s) = (1-2^{1-s}) \zeta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s} = -\displaystyle\lim_{z \to -1} \sum_{n=1}^\infty z^n n^{-s}$ where the last limit converges and is analytic for every $s \in \mathbb{C}$. It works because $\sum_{n=1}^N (-1)^{n+1} = \frac{1+ (-1)^{N+1}}{2}$ so the analytic continuation of $\eta(s)$ is obtained by partial summation – reuns Nov 16 '17 at 23:23
  • @reuns Can you point me to a reference which proves no such "natural topology" exists? – eepperly16 Nov 17 '17 at 01:19
  • Any book on complex analysis ? In general that $f_n \to f$ is a convergent sequence of analytic functions on $U$ thus converging uniformly, doesn't mean we can easily find (if it exists !) another sequence of analytic functions $g_n \to g$ on a larger open set, with $g = f$ on $U$. But when $f$ is entire, its Taylor series converge everywhere, thus it suffices to take $g_n$ the Taylor sums of $f$. – reuns Nov 17 '17 at 01:25
  • Do you see what I mean ? Here $f(s) = \zeta(s), f_n(s) = \sum_{m=1}^n m^{-s}, U = { \Re(s) > 1}$. – reuns Nov 17 '17 at 18:26
  • @reuns I'm sorry I don't understand why this implies there doesn't exist a new topology on $\mathbb C$ such that $\sum_{n=1}^\infty n^{-s}$ converges to $\zeta(s)$ for all $s$. – eepperly16 Nov 17 '17 at 18:50
  • I showed you some in my 1st comment... But it is specific to $\zeta(s)$. – reuns Nov 17 '17 at 18:54
  • @reuns Ok I think I get it. Thanks! – eepperly16 Nov 17 '17 at 19:18
  • May be useful: https://mathoverflow.net/questions/115743/an-algebra-of-integrals/342651#342651 – Anixx Dec 28 '20 at 22:11

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