Conjecture about divergent periodic summations of odd functions $$f(x)=\lim_{m\to\infty}\frac{\sum_{n=-m}^m(x-n)^{1/3}}{\sum_{n=-m}^m(1-n)^{1/3}}\stackrel{?}{=}x$$
I had a 'proof' but I made the simple mistake of assuming 2 limits could be swapped, so I only really proved that $f(x)=x$ for $x\in[-1,0,1]$, which is pretty trivial. Here's the 'proof' that I had. This was a first draft so it may be confusing to read.
The $(\text{stuff})^{1/3}$ in the numerator and denominator is really just $g(\text{stuff})$. Where $g$ just needs to be any odd function and relatively smooth and have a few other properties that I don't remember.
 A: We prove the following claim:

Claim. Suppose $g : \mathbb{R} \to \mathbb{R}$ satisfies the following condition:

*

*$g$ is an odd function.


*There exists $r \geq 0$ such that $g$ is either convex or concave on $[r, \infty)$.


*$\displaystyle \lim_{x\to\infty} |g(x)| = \infty$.


*For any $a \in \mathbb{R}$, $\displaystyle \lim_{x\to\infty} g(x+a)/g(x) = 1 $.
Then for any $x \in \mathbb{R}$,
$$ f(x) = \lim_{m\to\infty} \frac{\sum_{n=-m}^{m} g(x-n)}{\sum_{n=-m}^{m} g(1-n)} = x. $$

For example, $g(x) = \operatorname{sign}(x)|x|^p$ works for any $p > 0$, and this includes $g(x) = x^{1/3}$ as a special case.
Proof. By replacing $g$ by $-g$ if necessarily, we may assume that $g$ is convex on some $[r, \infty)$. also, it suffices to consider the case $x > 0$. Write
$$ G_m(x) = \sum_{n=-m}^{m} g(x-n). $$
Fix a positive integer $N$ such that $N > r + x + 1$. Then for $m \geq N$,
$$ G_m(x) = G_{N-1}(x) + \sum_{n=N}^{m} (g(n+x) - g(n-x)). $$
If $n \geq N$, then $g$ is convex on $[n-x-1, \infty) \subseteq [r, \infty)$, and so,
$$ g(n-x) - g(n-x-1) \leq \frac{g(n+x) - g(n-x)}{2x} \leq g(n+x+1) - g(n+x). $$
So it follows that
\begin{align*}
&G_{N-1}(x) + 2x(g(m-x) - g(N-x-1)) \\
&\quad\leq G_m(x) \\
&\qquad \leq G_{N-1}(x) + 2x(g(m+x+1) - g(N+x))
\end{align*}
Dividing both sides by $\sum_{n=-m}^{m} g(1-n) = g(m) + g(m+1)$ and letting $m\to\infty$, the desired conclusion follows by the Squeezing Theorem. $\square$

Remarks.

*

*The condition 3 that $|g(x)| \to \infty$ is necessary. Indeed, if we consider $g(x) = \arctan x$, then all of the conditions 1, 2, 4 are satisfied but we can prove that $f(x)$ exists but is not identical as $x$ (although it is very close to $x$).


*The condition 4 that $g(x+a)/g(x) \to 1$ is also necessary. Indeed, for $g(x) = \sinh x$ we have $f(x) = \sinh x/\sinh 1$.
