Consider this infinite series:
$1-\frac{1}{2^s} +\frac{1}{3^s} -\frac{1}{4^s} + ... $
I know that the alternate harmonic series converges to $\ln(2)$ - that will give me a value of $ s=1$. But how can I find the value of s if I don't know the above fact/property?
I tried the ratio test and got $$\lim_{n\to\infty} \left ( \frac{n}{n+1} \right )^{s} > -1$$
LHS approaches one - so does that mean it works for all values of s?
The alternating series theorem says the sum converges for all $s \gt 0$. For $s \le 0$ the terms do not go to zero so the sum does not converge.
The Dirichlet eta function, $\eta(s)$, is represented by the series $$\eta(s)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}=1-\frac{1}{2^s} +\frac{1}{3^s} -\frac{1}{4^s} + \dots$$
and if $s$ is purely real, then Dirichlet's test guarantees that the series representation converges for $\text{Re}(s)=s>0$ as shown inside this answer. On the other hand, if $\text{Re}(s)=s\le 0$, then the series diverges.
Following your approach, the ratio test is inconclusive when $s$ is finite since
$$L=\lim_{n\to\infty}\left|\frac{(-1)^{n+2}}{(n+1)^s}\frac{n^s}{(-1)^{n+1}}\right|=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^s=1$$
Therefore, more refined tests are required to determine convergence or divergence.
