There are many known telescoping series for $\zeta(s)$ and I was playing with the following two:
$$\displaystyle \zeta(s) = \frac{1}{(s-1)} \left(\sum _{n=1}^{\infty } \left( {\frac {n}{(n+1)^{s}}} - \frac{n-s}{n^s}\right) \right), \qquad \Re(s)>0$$
and
$$\displaystyle \zeta(s) = \frac{1}{(s-1)} \left(\sum _{n=1}^{\infty } \left( {\frac {n-1+s}{n^{s}}} - \frac{n-1}{(n-1)^s}\right) \right), \qquad 0<\Re(s)<1$$
After adding the two together and then dividing the result by $2$, we get:
$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(\sum _{n=1}^{\infty } \left( {\frac {n}{(n+1)^{s}}} + \frac{2\,s-1}{n^s} - {\frac {n-1}{\left( n-1 \right) ^{s}}}\right) \right), \qquad 0<\Re(s)<1$$
The numerator of the middle term is only dependent on $s$ and its real part becomes $0$ when $\Re(s)=\frac12$. This immediately yields a very simple expression for:
$$\displaystyle \zeta\left(\frac12\right) = \sum _{n=1}^{\infty } \left(\sqrt{n-1} -{\frac {n}{\sqrt{n+1}}}\right)\qquad(1)$$
Each individual term diverges, however assuming the RH is true, one could reason that it is the middle term (i.e. the only one with a full dependency in the numerator on $s$ and its real part becoming $0$ at $\Re(s)=\frac12$) that is responsible for precisely balancing the left and right series to become zero. The symmetry between the three series is not perfect and using the fact that $\displaystyle\sum _{n=1}^{\infty } \left( {\frac {1}{(n+1)^{s}}} - \frac{1}{n^s}\right)=-1$, it can be rewritten into a series that (imo) has a bit more "beauty":
$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} \left(1+\sum _{n=1}^{\infty } \left( {\frac {n+1}{(n+1)^{s}}} + \frac{2\,(s-1)}{n^s}-{\frac {n-1}{\left( n-1 \right) ^{s}}}\right) \right), \qquad -1<\Re(s)<1$$
which becomes:
$$\displaystyle \zeta(s) = \frac{1}{2\,(s-1)} + \sum _{n=1}^{\infty } \left( {\frac {(n+1)^{1-s}-(n-1)^{1-s}}{2\,(s-1)}} + \frac{1}{n^s}\right), \qquad -1<\Re(s)<1$$
$$\displaystyle \zeta\left(\frac12\right) = \sum _{n=1}^{\infty } \left(\sqrt{n-1} + \frac{1}{\sqrt{n}} -{ \sqrt{n+1}}\right)-1\qquad(2)$$
Are (1) and (2) known series for $\zeta\left(\frac12\right)$? (I could not find any reference to them on the web).