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.



proving inequality in Riemann zeta function
The inequality you want to prove is false. For example,
$$f\_5(1/2,1)=0.6096\dots,$$
while
$$\lim\_{k\to\infty}f\_k(1/2,1)=0.6398\dots$$


