Recently I have made some interesting observations on the limit $$\lim_{k\rightarrow \infty}{\sum_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}}. $$ When this limit exist denoteits convergence point by $\zeta$. If we define $f_k$ to be the partial sums of this series:
$$f_k(\alpha,\beta):= \sum_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}} $$ I want to prove that $$f_{2k}(\alpha,\beta)\le \zeta \le f_{2k+1}(\alpha,\beta)$$ for all values k. One approach is to prove them separately. First show $\zeta-f_{2k}(\alpha,\beta)\ge 0$ and then $f_{2k+1}(\alpha,\beta) - \zeta \le 0$.
From this we can easily obtain small bounds on the actual value of the zeta function and hopefully show that the there is only one value of $\alpha$ such that both the limits $$\lim_{k\rightarrow \infty}{\sum_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \cos \left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}}$$ and $$\lim_{k\rightarrow \infty}{\sum_{n=1}^{k}{\dfrac{(-1)^{n-1}\biggl( \sin\left(\beta\ln(n)\right)\biggr)}{n^{\alpha}}}}$$ are both simultaneously zero. We should be able to bound zeta by the first few terms.
