Zeta pushed to the limit 2

$\begin{aligned} \lim_{s \to \infty} \frac{\zeta(s) - 1}{\zeta(s - 1) - 1} &= \lim_{s \to \infty} \frac{\frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dotsb}{\frac{1}{2^{s - 1}} + \frac{1}{3^{s - 1}} + \frac{1}{4^{s - 1}} + \dotsb} \\ &= \lim_{s \to \infty} \frac{1 + \frac{2^s}{3^s} + \frac{2^s}{4^s} + \dotsb}{2 + \frac{2^s}{3^{s - 1}} + \frac{2^s}{4^{s - 1}} + \dotsb} \\ &= \frac{1}{2}. \end{aligned}$

We begin by taking the logarithm of the expression:

$\ln \left[ \lim_{s \to \infty} \frac{\zeta(s)-1}{\zeta(s-1)-1} \right] = \lim_{s \to \infty} \ln \left[ \frac{\zeta(s)-1}{\zeta(s-1)-1} \right] = \lim_{s \to \infty} \left[ \ln(\zeta(s)-1)-\ln(\zeta(s-1)-1) \right]$

We then multiply the whole thing by 1, in a smart way.

$\lim_{s \to \infty} \left[ \ln(\zeta(s)-1)-\ln(\zeta(s-1)-1) \right] = \lim_{s \to \infty} \left[ \frac{s}{s} \ln(\zeta(s)-1)-\frac{s-1}{s-1} \ln(\zeta(s-1)-1) \right]$

$= \lim_{s \to \infty} \left[ s \ln((\zeta(s)-1)^{1/s})-(s-1) \ln((\zeta(s-1)-1)^{1/s}) \right]$

Now, we use our result from the first problem, ie $\displaystyle \lim_{s \to \infty} (\zeta(s)-1)^{1/s} = \frac{1}{2}$ to further simply the expression.

$\lim_{s \to \infty} \left[ s \ln((\zeta(s)-1)^{1/s})-(s-1) \ln((\zeta(s-1)-1)^{1/s}) \right] = \lim_{s \to \infty} \left[ s \ln \frac{1}{2} - (s-1) \ln \frac{1}{2} \right] = \ln \frac{1}{2}$

Hence,

$\ln \left[ \lim_{s \to \infty} \frac{\zeta(s)-1}{\zeta(s-1)-1} \right] = \ln \frac{1}{2}$

and so

$\lim_{s \to \infty} \frac{\zeta(s)-1}{\zeta(s-1)-1} = \boxed{\frac{1}{2}}$