Zeta pushed to the limit 1

First, we can express $\zeta(s) - 1$ as the sum:

$\zeta(s) - 1$ = $\displaystyle \sum_{n=2}^\infty \frac{1}{n^{s}}$

Then, we state that, as $s$ goes to infinity, the following equality becomes true:

$\displaystyle \sum_{n=2}^\infty \frac{1}{n^{s}}$ = $\displaystyle \int_2^\infty \! \frac{1}{n^{s}} \, \mathrm{d}n.$

(I don't really know how to formally proof this, but if you compare the graph of the step function corresponding to the sum at the left side and the graph of $\frac{1}{n^{s}}$ , and take $s$ closer and closer to infinity, it's quite believable =p).

Solving the integral, we find:

$\lim_{s \to +\infty} \Bigg({ \displaystyle \int_2^\infty \! \frac{1}{n^{s}} \, \mathrm{d}n.}\Bigg)^\frac {1}{s}$ = $\lim_{s \to +\infty} \frac {n^{\frac {1}{s}}}{(s-1)^{\frac {1}{s}}n}$ , from infinity to two.

Since $\lim_{s \to +\infty} s^{\frac {1}{s}} = 1$ and $\lim_{s \to +\infty} (s-1)^{\frac {1}{s}} = 1$ , we have:

$\lim_{s \to +\infty} \frac {n^{\frac {1}{s}}}{(s-1)^{\frac {1}{s}}n}$ , from infinity to two = 0.5

$\begin{aligned} L & = \lim_{s \to \infty} (\zeta(s)-1)^\frac 1s \\ & = \lim_{s \to \infty} \left( 1 + \frac 1{2^s} + \frac 1{3^s} + \frac 1{4^s} + \cdots - 1\right)^\frac 1s \\ & = \lim_{s \to \infty} \left(\frac 1{2^s} + \frac 1{3^s} + \frac 1{4^s} + \cdots \right)^\frac 1s \\ & = \lim_{s \to \infty} \frac 12 \left(1 + \left(\frac 23 \right)^s + \left(\frac 24 \right)^s + \left(\frac 25 \right)^s + \cdots \right)^\frac 1s \\ & = \frac 12 = \boxed{0.5} \end{aligned}$

First, we express $\zeta(s)-1$ as a sum:

$\zeta(s)-1 = \sum_{n=2}^{\infty} \frac{1}{n^s}$

Now, we use that, $\displaystyle \lim_{N \to \infty} \left( \sum_{a \in A} a^N \right)^{1/N} = \max A$ when all elements of $A$ are positive, and thus evaluate the limit:

$\lim_{s \to \infty} (\zeta(s)-1)^{1/s} = \lim_{s \to \infty} \left( \sum_{n=2}^{\infty} \frac{1}{n^s} \right)^{1/s} = \max \lbrace \frac{1}{2}, \frac{1}{3}, \ldots \rbrace = \boxed{\frac{1}{2}}$