Zeta function summation Looking at the question here Surprising identities / equations shows an interesting identity, $$\large (\zeta (2)-1)-(\zeta (3)-1)+(\zeta(4)-1)-\cdots=\frac 12$$
How can we prove that result, or even prove that it is rational without proving that it equals $\frac 12 $?
 A: An alternative to changing the order of summation is to change a sum and an integral (no, it's not that different...). It is well-known that
$$ \zeta(s) = \frac{1}{\Gamma(s)} \int_0^{\infty} \frac{x^{s-1}}{e^x-1} \, dx, $$
which we can prove by expanding the denominator, using the binomial theorem on $e^{-x}(1-e^{-x})^{-1}$, which is perfectly fine since $|e^{-x}|<1$ for $x>0$; we then use the usual gamma integral. This immediately suggests a related formula for $\zeta(s)-1$: adding an $e^{-x}$ to the numerator just means we expand $e^{-2x}(1-e^{-x})^{-1}$, with the obvious result. Hence we can sum
$$ \sum_{k=2}^{\infty} (-1)^k(\zeta(k)-1) = \sum_{k=2}^{\infty} \frac{(-1)^k}{\Gamma(k)} \int_0^{\infty} \frac{x^{k-1}e^{-x}}{e^x-1} \, dx. $$
Interchanging the order of summation and integration, we arrive at
$$ \int_0^{\infty} \frac{e^{-2x}}{1-e^{-x}} \sum_{k=2}^{\infty} \frac{(-1)^k}{\Gamma(k)} x^{k-1} \, dx = \int_0^{\infty} \frac{e^{-2x}}{1-e^{-x}} (1-e^{-x}) \, dx \\
= \int_{0}^{\infty} e^{-2x} \, dx = \frac{1}{2}. $$

Edited to add:
This generalises nicely to sums of the form
$$ \sum_{k=2}^{\infty} \binom{k-1}{r} (\zeta(k)-1) = \zeta(r+1); $$
the alternating sums seem to lack such a simple expression, but can also be expressed in a similar way.
Edit 2:
Further work reveals
$$ \sum_{k=2}^{\infty} (-1)^k\binom{k-1}{r} (\zeta(k)-1) = \zeta(r+1)-1-2^{-r-1}. $$
A: We have that for any $s$,
$\zeta(s)-1=\sum_{n=2}^{\infty} \frac{1}{n^s}$.
Therefore we have
$\sum_{k=2}^{\infty} (-1)^k( \zeta(k)-1) = \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} (-1)^k \frac{1}{n^k} = \sum_{n=2}^{\infty} \sum_{k=2}^{\infty} (-\frac{1}{n})^k = \sum_{n=2}^{\infty} \frac{\frac{1}{n^2}}{1+\frac{1}{n}} = \sum_{n=2}^{\infty} \frac{1}{n(n+1)} = \sum_{n=2}^{\infty} \frac{1}{n}-\frac{1}{n+1} = \frac{1}{2}.$
Now before someone yells at me, we need to justify the interchange of summations in that second equality there. I believe this can be accomplished by applying Fubini's Theorem via counting measure (although someone may want to recheck that).
A: Here is the basic idea: reordering terms
$$\eqalign{LHS
  &=\sum_{n=2}^\infty\frac{1}{n^2}-\sum_{n=2}^\infty\frac{1}{n^3}
    +\sum_{n=2}^\infty\frac{1}{n^4}-\cdots\cr
  &=\Bigl(\frac1{2^2}-\frac1{2^3}+\frac1{2^4}-\cdots\Bigr)+
    \Bigl(\frac1{3^2}-\frac1{3^3}+\frac1{3^4}-\cdots\Bigr)+\cdots\cr
  &=\sum_{k=2}^\infty\frac1{k^2}\frac{k}{k+1}\cr
  &=\sum_{k=2}^\infty\Bigl(\frac1k-\frac1{k+1}\Bigr)\cr
  &=\frac12\ .\cr}$$
To give a proper proof you need to show that the reordering is permissible.
