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By analyzing some patterns that Wolfram generated, I believe this identity is true. $$\sum_{i=1}^m\frac{1}{i^s}=\sum_{n=0}^\infty\left(\frac{(-1)^n}{n!}\sum_{i=1}^m{\ln^n(i)}\right)s^n$$

Since this is the Riemann zeta function when we limit $m$ to infinity for $\Re(s)\gt 1$, what I did next was try to come up with a similar representation for the Riemann zeta series with expansion to $0\le\Re(s)\le 1$.

Once again I looked at patterns and so I believe that $$\frac{1}{1-2^{1-s}}\sum_{i=1}^m\frac{(-1)^{n}}{i^s}=\sum_{n=0}^\infty\left((-1)^n\sum_{k=0}^n(-1)^k\frac{A(k)}{(n-k)!}\ln^k(2)\sum_{i=1}^m(-1)^i\ln^{n-k}(i)\right)s^n$$ where $$A(x)=\prod_{i=0}^x\frac{\left[\frac{i!}{\ln(2)^{i+1}}\right]}{\left[\frac{i!}{\ln(2)^i}\right]}$$ (The brackets mean round to the nearest integer.) $$\\$$ How can we prove the first and second equations mathematically?

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    I dont know how to derive it but you can try to prove that your identities are true through induction on $m$. – 2017-02-28

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It should be rather clear for the first one that

$$\sum_{n=0}^\infty\frac{(-1)^n}{n!}\sum_{i=1}^m\ln^n(i)s^n=\sum_{i=1}^m\sum_{n=0}^\infty\frac{(-s\ln(i))^n}{n!}=\sum_{i=1}^me^{-s\ln(i)}=\sum_{i=1}^mi^{-s}$$

While the second looks like a horrendous remake of the first.