Weird thing about $\sum_{k=2}^{\infty}(-1)^k\zeta{(k)}$

Question

Consider the sum $S=\sum_{k=2}^{\infty}(-1)^k\zeta{(k)}$. By a simple manipulation, we can show: $$ S=\sum_{k=2}^{\infty}{(-1)^k\sum_{r=1}^{\infty}\frac{1}{r^k}}=\sum_{r=1}^{\infty}{\sum_{k=2}^{\infty}\frac{(-1)^k}{r^k}}=\sum_{r=1}^{\infty}{\frac{1}{1+\frac{1}{r}}-1+\frac{1}{r}}=\sum_{r=1}^{\infty}{\frac{1}{r}-\frac{1}{r+1}}=1 $$ But looking at the partial sums $S_n$ we can see that: $$ \begin{array}{c|lcr} n & S_n\\ \hline 2 & 1.644934... \\ 3 & 0.442877... \\ 4 & 1.525200... \\ 100 & 1.500…026… \\ 101 & 0.499…987... \end{array} $$ Which disagrees with the result. How can one be trapped so easily?

Edit:

And how to show $$ \lim_{n\to\infty}{S_n-\frac{(-1)^n+1}{2}}=\frac{1}{2} $$ ?

Answer 1

Hint. The series is not convergent since $$ \lim_{k \to \infty}(-1)^k\zeta(k)\neq 0. $$ One may recall that $$ \zeta(k)=\color{red}{1}+\frac1{2^k}+\frac1{3^k}+\cdots. $$

Answer 2

What is convergent is $$\displaystyle \sum_{i \geq 2} (-1)^i( \zeta(i)-1)=\sum_{k \geq 2}\left( \sum_{i \geq 2} {1 \over (-k)^i} \right)=\sum_{k \geq 2} {1 \over k(k+1)}=\sum_{k \geq 2} {1 \over k(k+1)}=\sum_{k \geq 2} {1 \over k}-{1 \over k+1}= \lim_{n \to \infty}{1 \over 2}-{1 \over n+1}={1 \over 2}$$

Yours of course doesn't converge, because the summed terms don't even go to $0$. However, you can easily prove your limit because: $${1 \over 2}=\sum_{i \geq 2} (-1)^i( \zeta(i)-1)= \lim_{n \to \infty} S_n- \sum_{i=2}^{n}(-1)^i=\lim_{n \to \infty} S_n - {(-1)^n+1 \over 2}$$