How to prove this infinite series identity? Prove that:
$$
1-\frac{1}{2^{n}}+ \frac{1}{3^{n}} - \frac{1}{4^{n}} + \cdots = \left(1-\frac{1}{2^{n-1}}\right)\zeta(n)
$$
where $n>1$ and:
$$
\zeta(n) = 1+\frac{1}{2^{n}}+ \frac{1}{3^{n}} + \frac{1}{4^{n}} + \cdots
$$
I have verified this is true using numerical method, but how to get the exact proof?

Hint:
Rewrite the sum as
$$1-\frac{1}{2^{n}}+ \frac{1}{3^{n}} - \frac{1}{4^{n}} + \cdots=\zeta(n)-2\Bigl(\frac{1}{2^{n}}+ \frac{1}{4^{n}} + \cdots\Bigr)=\zeta(n)-\frac{2}{2^n}\Bigl(1+ \frac{1}{2^{n}} + \cdots\Bigr)$$
Can you end the computation?
