How can we write $\zeta(n)(2^{n}-1)$ for $n \in \mathbb{N}$ and $n \geq 1$ in other form?
I wrote the term as $$2^n\left(1-\frac{1}{2^n}\right)\zeta(n)= 2^n\frac{1}{\sum_{r=0}^\infty \frac{1}{2^{nr}}}\sum_{r=0}^\infty\frac{1}{(r+1)^n}$$
After this step, I don't know how shall I proceed. Any ideas/suggestions are highly appreciated.
Since $\zeta(n) = 1 + 1/2^n + 1/3^n + \cdots $, we know that for $n>1$ the series converges absolutely therefore we can rearrange it without changing the limit value. So $$(1- \frac 1{2^n})(1+ \frac 1{2^n} + \frac 1{3^n} + \cdots ) = 1+ \frac 1{2^n} + \frac 1{3^n} + \cdots - ( \frac 1{2^n} + \frac 1 {4^n} + \frac 1 {6^n} + \cdots ) = 1 + \frac 1 {3^n} + \frac 1 {5^n} + \cdots $$
So we have $\zeta(n)(2^n -1 ) = \sum_{k = 1}^{\infty} \frac{2^n}{(2k+1)^n} = \sum_{k = 1}^{\infty} \frac{1}{(k+\frac 12)^n}$