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.
(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 )
