Is it true that if $\lim_{s \to 0} \sum_{n=0}^\infty \frac{a_n}{s^{n+1}}$ converges, then $a_n = (-1)^n a_0$? Is it true that if $\lim_{s \to 0} \sum_{n=0}^\infty \frac{a_n}{s^{n+1}}$ converges, then $a_n = (-1)^n a_0$? The original problem was to prove the coefficients of a entire holomorphic function is of this form, which I used laplacian transform and get this equivalent condition. My attempt is quite non-rigorous, which is to use symmetry and let $$  \sum_{n=0}^\infty \frac{a_n}{s^{n+1}} = \frac{a_0 + \sum_{n=1}^\infty \frac{a_n}{s^{n}}}{s}$$
and because $\sum_{n=1}^\infty \frac{a_n}{s^{n}} =  s\sum_{n=0}^\infty \frac{a_n}{s^{n+1}}- a_0$ is convergent at $0$, we can infer that $a_0 = -\lim_{s \to 0}\sum_{n=1}^\infty \frac{a_n}{s^{n}}$.
WLOG let $a_0 = 1$, and let $S_1$ denote the possible solutions for $a_1, a_2, \dots,a_n,\dots$ such that $1 = -\lim_{s \to 0}\sum_{n=1}^\infty \frac{a_n}{s^{n}}$. We choose one of them in which $a_1 = \lambda$. Now we fix $a_1 = \lambda$, note that by symmetry suppose $a_1 = -\lim_{s \to 0}\sum_{n=1}^\infty \frac{a_{n+1}}{s^{n}}$ and thus $1 = -\lim_{s \to 0}\sum_{n=1}^\infty \frac{\frac{a_{n+1}}{a_1}}{s^{n}}$, then the solution for this equation will also be the solution for the previous one. But note this is the exactly same equation above. So we choose the same solution in $S_1$ such that $\frac{a_{2}}{a_1}= \lambda$. Thus we have that if $a_1 = \lambda$, then $a_n = \lambda^n$ will be one of solutions for the original equation. Empiricial computation show that only $\lambda = -1$, $\lim_{s \to 0} \sum_{n=0}^\infty \frac{\lambda^n}{s^{n+1}}$ will converge. Thus the statement is true. Is this reasoning correct?
 A: If $\ \displaystyle\lim_{s \to 0} \sum_{n=0}^\infty \frac{a_n}{s^{n+1}}\ $ converges, then $\ \displaystyle\lim_{s \to 0^+} \sum_{n=0}^\infty \frac{a_n}{s^{n+1}}\ $ converges, i.e. $\ \displaystyle\lim_{t \to\infty} \sum_{n=0}^\infty a_nt^{n+1}\ $ converges,
which implies that $\ a_n\to 0\ $ as $\ n\to\infty.$
If there are non-zero terms, but there is no final non-zero term in our sequence $(a_n)$, i.e. there is no $\ k\in\mathbb{N}\cup\{0\}\ $ such that $\ a_n=0\ \forall n\geq k,\ $ then we can always find a $\ t\ $ large enough so that the absolute value of arbitrarily many of the the non-zero terms in our series is greater than $1$. This shows our series will not converge as $\ t\to\infty\ $ if there are non-zero terms but no final non-zero term in our sequence $(a_n).$
Alternatively, if there is a final non-zero term in the sequence, then clearly our series will not converge because the last non-zero term will dominate as $t\to \infty.$
Therefore, the only possible sequence $(a_n)$ can be is the all-zero sequence:
$a_n = 0\ \forall n\in\mathbb{N}\cup\{0\}.$
[Note that this sequence also gives rise to a convergent series as $\ t\to-\infty\ $ , as required.]
This means that crazy attempts of making the series converge, like $\ a_n = \Large{\frac{1}{2^{2^n}}}\ $ fail as $\ t\to\infty.$
It also means your proposition:

If $\lim_{s \to 0} \sum_{n=0}^\infty \frac{a_n}{s^{n+1}}$ converges,
then $a_n = (-1)^n a_0$

is technically true.
