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Let $\Sigma_\text{p} : (\mathbb{N} \to \mathbb{C}) \rightharpoonup \mathbb{C}$ where $\Sigma_\text{p}(a)$ is the analytic continuation of the power series $\sum_{n=0}^\infty a_n x^n$ to $x=1$, when the latter exists. Examples: \begin{array}{l|l|l|l} a_n & a & f(x) & f(1) = \Sigma_\text{p}(a) \\ \hline 2^n & 1, 2, 4, 8, 16, \ldots & \frac{1}{1-2x} & -1 \\ 3^n (n+1) & 1, 6, 27, 108, 405, \ldots & \frac{1}{(1-3x)^2} & \frac{1}{4} \\ \binom{2n}{n} & 1, 2, 6, 20, 70, \ldots & \frac{1}{\sqrt{1-4x}} & -\frac{i}{\sqrt{3}} \\ \text{Fibonacci}_n & 0, 1, 1, 2, 3, \ldots & \frac{x}{1-x-x^2} & -1 \\ \text{Motzkin}_n & 1, 1, 2, 4, 9, \ldots & \frac{1 - x - \sqrt{1 - 2x - 3x^2}}{2x^2} & -i \end{array}

$\sum_{n=0}^\infty a_n = \Sigma_\text{p}(a)$ when the LHS exists. Thus we can extend the definition of summation to some divergent series by defining the LHS as the RHS.

Under this definition, sequences of natural numbers can "sum" to negative, fractional, irrational, and even imaginary numbers. Are there "simple" (ideally, closed-form) sequences of natural numbers that sum to the following constants?

If necessary, analytic continuation of other series like Dirichlet series or general Dirichlet series may be used. For example, let $\Sigma_\text{D} : (\mathbb{N} \to \mathbb{C}) \rightharpoonup \mathbb{C}$ where $\Sigma_\text{D}(a)$ is the analytic continuation of the Dirichlet series $\sum_{n=0}^\infty a_n n^{-x}$ to $x=0$, when the latter exists. (See this question.)

user76284
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  • If you can express $a$ and $b$ in that way, then you can also express $ab$. Just multiply the sequences as they were polynomials with same coefficients. If you add the sequences you get $a+b$. You basically have a method for getting any rational number. If you found a method to get square roots, you would answer the case of first 3 constants you gave for examples. – cnikbesku Dec 13 '22 at 22:36
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    Why is $-\frac i{\sqrt3}$ the right answer in the last example instead of $+\frac i{\sqrt3}$? The expression $\sqrt{-1}$ is not canonically equal to $i$. Indeed, the fact that the sign in this example depends on exactly how one analytically continues the power series to $x=1$ (a branch cut is needed at $x=\frac12$), which to me illustrates that these regularizations should not in any way be considered similar to convergent series. – Greg Martin Dec 13 '22 at 23:43
  • Wait a second! There is no the analytic continuation. Yeah, just like @GregMartin said, you can get branches. It is an interesting topic to discuss holomorphic functions whose series at 0 consist of all natural numbers. But we would have to give a better definition of a valid representation. – cnikbesku Dec 14 '22 at 01:27
  • @donaastor Do you mean meromorphic? – user76284 Dec 14 '22 at 15:53
  • Yeah, I meant that globally, but I titled the topic with holomorphic because it is kinda holomorphic in the neighborhood of 0. We you start the analytic continuation we should switch to kind of meromorphic term. But in general, maybe we don't have to require it to be globally meromorphic, that's why I am saying let's agree on a formal definition. Btw, I did find a way to represent any square root of whole numbers, but the problem might become that branching. So the first 3 constants should be possible. Oops, all square roots except for $\sqrt{2}$... – cnikbesku Dec 14 '22 at 15:55
  • @donaastor What's the way? – user76284 Dec 14 '22 at 16:02
  • In the third construction of yours, take $\binom{2n}{n}k^n$ instead and you will get to evaluate the same function but instead of at $x=1$, you evaluate at $x=k$, so you get $\frac{-i}{\sqrt{4k-1}}$. Now for every $n\equiv_4 3$ you can easily get $\frac{-i}{\sqrt{n}}$ and for $n\equiv_4 1$ you can get $\frac{-i}{\sqrt{3n}}$. Multiply the latter by $\frac{-i}{\sqrt{3}}$ then by $-3$ and you get $\frac{1}{\sqrt{n}}$ for every $n\equiv_4 1$. Now take $\frac{-i}{\sqrt{5n}}$ for $n\equiv_4 3$ and multiply by $\frac{1}{\sqrt{5}}$ to get $\frac{1}{\sqrt{n}}$ for all odd $n$. Mult. by $n$ to get all. – cnikbesku Dec 14 '22 at 16:11
  • Sort of, I wrote $\frac{1}{\sqrt{5}}$ instead of $\frac{-i}{\sqrt{5}}$. And then you would multiply by $-1$ agian. But you get the trick... – cnikbesku Dec 14 '22 at 16:27

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