Analytic continuation for $\zeta(-x)$ vs $\frac{1}{(1-x)^2}$ at $x=1$

Question

Everyone and their grandma has seen youtube videos about the analytic continuation of the series definition of the zeta function. My question is regarding another function whose series expansion seems to "evaluate" to 1+2+3+..., namely $\frac{1}{(1-x)^2}$ at $x=1$. This function has a genuine second order pole at the point where the terms of its series expansion evaluates to 1+2+3+...

1- Consider $F(x)=\sum f_n(x)$ which converges to an analytic function in some region of the complex plane. Furthermore, at some point $a$, which is outside of the region of convergence, we get $f_n(a)=n$. For example, $\sum n^{-x}$ and $\sum nx^{n-1}$ are both such series. Is it possible to come up with an infinite variety of such series which, once analytically continued, can yield any complex value: -1/12, complex infinity, anything in between.

2- Is there a reason to privilege the zeta function over $\frac{1}{(1-x)^2}$ or other series whose terms evaluate to 1+2+3+... outside of its region of convergence? If one obtained an asymptotic series solution to a problem as $x \rightarrow +\infty$ that had the form $\sum n^{-x}$, of course the correct evaluation at $x=-1$ would really be $-\frac{1}{12}$, but is there any other reason to privilage the zeta function?

Answer 1

What your looking for, is basically a function whose series looks like it should evaluate to $0+0+\dots$, yet instead equals a finite value.

Consider $\sum (1-x)x^n$. The analytical continuation of this sum is of course just $\sum (1-x)x^n = 1$. However, at $x=1$, this sum looks like $\sum (1-1)(1) = 0 + 0 + \dots $.

Using this idea, we have by analytical continuation, that $$f(x) = \alpha \left(1-\left(\frac{1+x}{2}\right)\right)+\sum_{n=1}^{\infty}\left(n^{x}+\alpha\left(1-\left(\frac{1+x}{2}\right)\right)\left(\frac{1+x}{2}\right)^{n}\right) = \alpha + \zeta(-z)$$

However, at $1$ we have that $f(1) = \sum_{n=1}^\infty n = \alpha + \zeta(-z)$. So an arbitrary value can be achieved. This addresses (1).

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To address (2), $-\frac{1}{12}$ and $\infty$ are the only natural values to assign to this series. We can obtain the value of $-\frac{1}{12}$ in various ways which don't rely on analytical continuation, and consider only analyzing the sum $\sum n$. Many of these methods essentially rely on finding the constant term in the (smoothed) expansion of the partial sum at infinity. You can find various expositions of these methods in other answers on this site.

Of course, saying that $-\frac{1}{12}$ is the only natural answer seems to contradict (1). However, the ability to obtain contradictory results from the sum $\sum n$ isn't an indication that divergent series is broken, but rather that there are different rules for dealing with divergent series than regular series. In essence, the example provided above takes advantage of the fact that, for divergent series (and more generally, non-absolutely convergent series), associativity doesn't hold.

I basically used the fact that $(1-1)+(1-1)+(1-1)+\dots \neq 1+(-1+1)+(-1+1) + \dots$ to produce my example. In essence, I used the fact that $1 = 1 + (-x+x) = 1+(-x+x)+(-x^2+x^2) + \dots$ and rearranged this to $(1-x) + (x-x^2) + (x^2 -x^3) + \dots$ to obtain an incorrect regularization.

There is a correct way to handle divergent series, but treating them as regular convergent series doesn't work. A good place to start on how to handle them correct is here: https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation

Answer 2

If $x=e^{-y}$, then as $y\to0^+$ from the right half plane, we have

$$ \sum_{n\ge1}nx^n={1\over y^2}-{1\over12}+O(y). $$

However, if we remove the 2nd order pole from the right hand side, we see that $-{1\over12}$ is reasonably assigned to the divergent sum $\sum_{n\ge1}n$. More generally, if we let

$$ f(y;s)=\sum_{n\ge1}n^{-s}e^{-ny},\tag1 $$

where $\Re(s)=\sigma<1$. Then, it can be proven that when $|y|<2\pi$, we have the following absolutely convergent series representation:

$$ f(y;s)=\sum_{n\ge1}n^{-s}e^{-ny}=y^{s-1}\Gamma(1-s)+\zeta(s)+\sum_{n\ge1}{(-y)^n\over n!}\zeta(s-n).\tag2 $$

When $s\notin\mathbb Z$, we also need to remove the negative real axis to settle done issues with $y^{s-1}$.

Therefore, the expansion of (1) near $y=0$ is indeed helpful for us to obtain information about the Riemann zeta function.

Proof sketch of (2)

By the principle of analytic continuation, it is sufficient to assume $0<y<2\pi$, so we have

$$ f(y;s)=\sum_{n\ge1}n^{-s}{1\over2\pi i}\int_{2-\sigma-i\infty}^{2-\sigma+i\infty}y^{-z}\Gamma(z)\zeta(s+z)\mathrm dz. $$

Computing residues, we have

$$ \mathop{\operatorname{Res}}_{s=1-z}=y^{s-1}\Gamma(1-s), $$

$$ \mathop{\operatorname{Res}}_{s=0}=\zeta(s), $$

and when $n$ is a positive integer, we have

$$ \mathop{\operatorname{Res}}_{s=-n}={(-y)^n\over n!}\zeta(s-n). $$

Therefore, we obtain

$$ f(y;s)=y^{s-1}\Gamma(1-s)+\zeta(s)+\sum_{n=1}^N{(-y)^n\over n!}\zeta(s-n)+J_N,\tag3 $$

where

$$ J_N={1\over2\pi i}\int_{-N-\frac12-i\infty}^{-N-\frac12+i\infty}y^{-z}\Gamma(z)\zeta(s+z)\mathrm dz. $$

By applying the functional equation for $\zeta$ and Stirling's approximation for $\Gamma$. we deduce that when $z=-N-\frac12+iv$, there is

$$ |y^{-z}\Gamma(z)\zeta(s+z)|\ll(1-\sigma+N)^{\frac12-\sigma+N}\left(y\over2\pi\right)^N(N^2+v^2)^{-{N+1\over2}}, $$

It can be proven that

$$ \int_0^{+\infty}{\mathrm dv\over(N^2+v^2)^{N+1\over2}}=O(N^{-N-\frac12}), $$

so we have

$$ |J_N|\ll\left(1+{1-\sigma\over N}\right)^{N+\frac12}N^{-\sigma}\left(y\over2\pi\right)^N\ll N^{-\sigma}\left(y\over2\pi\right)^N. $$

Plugging this into (3) and letting $N\to+\infty$ give us (2).