## Background

Taking a relatively arbitrary combination of exponential and polynomial terms, for instance $$\sum_{n=0}^\infty \left(n^{2}\sin\left(n\right)+n\cos\left(3n-2\right)\right)\cos\left(5n+1\right)x^{n}$$ one will find that this series is 'miraculously' analytically continued by the same series except with $n$ replaced by $-n$ (additionally, with an extra negative sign and the index starting at $n=1$). In particular, the first series has an analytical continuation for $|x|>1$ given by $$-\sum_{n=1}^{\infty}\left(\left(-n\right)^{2}\sin\left(-n\right)-n\cos\left(3\left(-n\right)-2\right)\right)\cos\left(5(-n)+1\right)x^{-n}$$

In general, when $P(n) = \sum_{k=0}^N a_k n^k$ is a polynomial, then $\sum_{n=0}^\infty P(n)x^n$ is analytically continued by $- \sum_{n=1}^\infty P(-n)x^{-n}$. Likewise, when $F(n) = \sum_{k=0}^N a_k e^{i \theta_k n}$ is a Fourier-like series, then $\sum_{n=0}^\infty F(n) x^n$ is continued to $-\sum_{n=1}^\infty F(-n)x^{-n}$. The second result can be generalized to when $F(n)$ is almost periodic, and I imagine a similar result can be found when $P(n)$ is 'almost-polynomial' under a suitable definition.

One potential heuristic for why this relationship holds can be found by considering the residue theorem. If $(-1)^n f(n)$ is an entire function under sufficient growth conditions, then $\frac{1}{2 i} \int_{-1/2 - i\infty}^{-1/2 + i \infty} f(n) \csc(\pi n)$ is equal to both $\sum_{n=0}^\infty (-1)^n f(n)$ and $-\sum_{n=1}^\infty (-1)^n f(-n)$ by considering either a large rectangle covering everything to the right of the contour, or one covering everything to the left (the negative sign comes from switching the orientation of the contour).

This heuristic doesn't tell the whole story however. Consider the function $\frac{d}{dx}\frac{\Gamma'(x+1)}{\Gamma(x+1)}$ and its corresponding Taylor series $\sum_{k=0}^\infty (k+1) \zeta(k+2)(-x)^k$. $\zeta$ grows too large in the negative direction, so the contour does not converge and we can't support changing the direction of the contour. Nonetheless, it turns out that the series $$-\sum_{k=1}^\infty (-k+1) \zeta(-k+2)(-x)^{-k}$$ (where we understand $0\cdot \zeta(1)=1$) asymptotically extends the original series. In particular, if we fix $N$ then $\lim_{x \to +\infty} \sum_{k=1}^N (-k+1)\zeta(-k+2)(-x)^{-k} = \frac{d}{dx} \frac{\Gamma'(x+1)}{\Gamma(x+1)}$. In fact, taking only the first 10 terms of the sum gives an approximation where the difference between the series and the function at $x=3$ is less than $10^{-6}$ (and decreases to zero as $x$ becomes large).

Another case occurs with functions like $\sum_{n=0}^\infty \sin(\sin(n))x^n$, which is continued by $-\sum_{n=1}^\infty \sin(\sin(-n))x^{-n}$. In this case, analyzing the relationship between the two functions using the residue theorem seems even less clear, since $\sin(\sin(z))$ only remains small when $z$ has a small imaginary part.

Finally, I should note that its clear that there are cases where $\sum_{n=0}^\infty (-1)^nf(n)x^n \neq -\sum_{n=1}^\infty (-1)^nf(-n)x^{-n}$. For instance, if we choose $f(n) = \frac{1}{n^2+1}$, then $f(n) =f(-n)$. However, in simple cases like these we can salvage the relationship using the residue theorem to pick up the extra poles off the real line, so that, for example $$\sum_{n=0}^\infty \frac{(-1)^n x^n}{n^2+1} = -\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}x^{-n}}{n^{2}+1}+\pi\cos\left(\ln\left(x\right)\right)\operatorname{csch}\left(\pi\right)$$

## Question

In light of these examples, I am curious about general characterizations on the relationship between $F(x)=\sum_{n=0}^\infty f(n)x^n$ and $F_2(x)=-\sum_{n=1}^\infty f(-n) x^{-n}$. Some questions that come to mind are

- How are these sums connected when $f(n)$ is an entire function, but diverges fast enough at $\pm i \infty$ so that the residue theorem can't be applied?
- Are there other broad classes of functions where $F(x)$ is analytically continued to $F_2(x)$? Are there broad classes of functions where the relationship fails?
- How are the sums related when $F(x)$ is a divergent series, for instance $f(n)=(\alpha n)!$ or $f(n) = (n^2)^n$.
- Are there cases where $F(x)$ and $F_2(x)$ are related in some way, but aren't analytical continuations of each other?
- Is there uniqueness, in the sense that if $\sum_{n=0}^\infty f(n) x^n = \sum_{n=0}^\infty g(n) x^n$, must we also have $-\sum_{n=1}^\infty f(-n)x^{-n} = -\sum_{n=1}^\infty g(-n)x^{-n}$

## Edit: The Residue Approach

I've added this section to give a little bit more detail about what I am thinking about with the residue approach. In connection with Alexandre's answer, it is interesting to me that this approach seems to stop working at precisely the same point where $f(n)$ is no longer uniquely determined by by its values on the integers (by an application of Carlson's theorem).

We can apply the residue theorem to analyze sums of the form $\sum_{n=0}^\infty (-1)^n f(n)x^n$ where $x>0$ and and $f$ is an entire function of exponential type less than $\pi$ (i.e. $|f(r e^{i \theta})| < Me^{\pi r}$). With the latter constraint on the growth rate of $f$, through some algebriac manipulation we can obtain for the right half plane $\lim_{|z| \to \infty} f(z)x^z csc(\pi z) = 0$ when $x \leq \frac{1}{e^{\pi}}$. When $f(z)$ is an entire function, then, we have that $$\sum_{n=0}^\infty (-1)^n f(n)x^n = \frac{1}{2i}\int_C \csc(\pi z) f(z) x^z dz$$ Where $C$ is a rectangular contour whose height and width go to infinity. However, all but the contour at $Re(z) = -\frac{1}{2}$ go to zero by the growth conditions, and so $\sum_{n=0}^\infty (-1)^n f(n) x^n = \frac{1}{2i}\int_{-\frac{1}{2}- i \infty}^{-\frac{1}{2}+ i \infty} f(z)x^z \csc(\pi z)dz$ for $x< \frac{1}{e^{\pi}}$. However, as long as the integral is analytic, we now have two analytic fucntions whose agree on a set with a limit point, so they represent the same analytic function. Moreover, the contour continues to converge even when the original sum doesn't, so the contour integral analytically continues the original function. Moreover, when $x> e^{\pi}$ then the $f(z) \csc(\pi z)x^z$ is small for values on the left-half plane. So, we obtain $$\frac{1}{2i} \int_{-\frac{1}{2}- i \infty}^{-\frac{1}{2}+ i \infty} f(z)x^z \csc(\pi z)dz = \frac{1}{2i} \int_{C_2} f(z)x^z \csc(\pi z)dz = -\sum_{n=1}^\infty (-1)^nf(-n)x^{-n} $$ Since the other pieces of the contour $C_2$ are zero (i.e. all the edges except the one where $Re(z)=\frac{1}{2}$). $C_2$ is a rectangle of infinite width and height facing in the opposite direction to $C$ given by this picture

In the case where $f(n)$ is not entire, then we can usually shift the original contour to the line $Re(z) = c$ so that all the poles of $f$ are to the left of $c$. Then we obtain that, for $x< \frac{1}{e^{\pi}}$

$$\sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c+C_{\rightarrow}} f(z) x^z \csc(\pi z) dz = \\ \sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c-i \infty}^{c + i \infty} f(z) x^z \csc(\pi z) dz=\sum_{n=0}^\infty (-1)^n f(n) x^n$$
where $c+C_{\rightarrow}$ is the rectangular contour that covers everything to the right of $Re(z) = c$. And again for $x>e^{\pi}$ we can switch the direction of the contour, and this gives us that
$$\sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c+C_{\leftarrow}} f(z) x^z \csc(\pi z) dz = \\ \sum_{0 \leq n < c} (-1)^n f(n) x^n + \frac{1}{2i}\int_{c-i \infty}^{c + i \infty} f(z) x^z \csc(\pi z) dz=\\-\sum_{n=1}^\infty (-1)^n f(-n) x^{-n} + \frac{1}{2i} \sum_{} \text{Res}(f(z)\csc(\pi z)x^z, a_k)$$
Conviently, the extra terms $\sum_{0 \leq n < c} (-1)^n f(n) x^n$ that were added in initially cancel out since the $C_{\leftarrow}$ contour traverses those same points in the opposite direction.

Math. Zeit.29(1928–1929), 549–640. $\endgroup$4more comments