alternating sum of zeta functions minus one is one half During my work on a different infinite series I happened to prove that 
$\displaystyle\sum_{k=2}^{\infty}(-1)^k (\zeta(k)-1)=\frac{1}{2}$
 where
 $ \displaystyle\zeta(k)=\sum_{n=1}^{\infty}\frac{1}{n^k}$
My question is whether this relation was known before and is of any interest to others.  
 A: (Usually, the following were a comment, but since you're asking whether they are of interest... I'll make it an "answer". It is from an answer of mine to a thread "surprising identities" earlier in MSE)
Some zeta-identies have been much surprising to me.
Let's denote the value $\zeta(s)-1$ as $\zeta_1(s)$ then
$$ \small \begin{array} {}
1 \zeta_1(2) &+&1 \zeta_1(3)&+&1 \zeta_1(4)&+&1 \zeta_1(5)&+&  ... &=&1\\
1 \zeta_1(2) &+&2 \zeta_1(3)&+&3 \zeta_1(4)&+&4 \zeta_1(5)&+&  ... &=&\zeta(2)\\
             & &1 \zeta_1(3)&+&3 \zeta_1(4)&+&6 \zeta_1(5)&+&  ... &=&\zeta(3)\\ 
             & &            & &1 \zeta_1(4)&+&4 \zeta_1(5)&+&  ... &=&\zeta(4)\\ 
             & &            & &            & &1 \zeta_1(5)&+&  ... &=&\zeta(5)\\ 
 ...         & & & & & & & &... &= &  ...
\end{array}
$$
There are very similar stunning alternating-series relations:
$$ \small \begin{array} {}
1 \zeta_1(2) &-&1 \zeta_1(3)&+&1 \zeta_1(4)&-&1 \zeta_1(5)&+&  ... &=&1/2\\
             & &2 \zeta_1(3)&-&3 \zeta_1(4)&+&4 \zeta_1(5)&-&  ... &=&1/4\\
             & &            & &3 \zeta_1(4)&-&6 \zeta_1(5)&+&  ... &=&1/8\\ 
             & &            & &            & &4 \zeta_1(5)&-&  ... &=&1/16\\ 
 ...         & & & & & & & &... &= &  ...
\end{array}
$$
(An even older, likely difficult to read but more involved, discussion of identities like this and how one can arrive at them can be found in this pdf )
