I'm looking at Eqn. 13 here:
$$\sum (1/(2n-1)^s - 1/(2n)^s + 2/(2n)^s)$$
They then simplify this to obtain the zeta function:
$$\sum 1/n^s = \zeta(s)$$
Can you help with the simplication? All I can write is
$$1/(2n-1)^s + 1/(2n)^s$$
I don't know how to simplify this further. I tried Geogebra and Wolfram Alfa they both give different answers but they don't simplify to the zeta function.
EDIT: Wolfram alpha screen shot:
$$ \sum_{n\ge 1} [\frac{1}{(2n-1)^s} -\frac{1}{(2n)^s}+\frac{2}{(2n)^s}] = \sum_{n\ge 1} [\frac{1}{(2n-1)^s} +\frac{1}{(2n)^s}] $$ $$ = \sum_{n\ge 1} \frac{1}{(2n-1)^s} +\sum_{n\ge 1} \frac{1}{(2n)^s} $$ $$ = \sum_{n=1,3,5,7,\ldots odd} \frac{1}{n^s} +\sum_{n=2,4,6,8,10 \ldots even} \frac{1}{n^s} $$ $$ = \sum_{n=1,2,3,4,5,} \frac{1}{n^s} = \zeta(s). $$