I wrote some similar questions:
I wanted to solve the problem myself but i must ask for the last time. This time this is general question.
Let: $$A_{n}(x)=\sum_{k=1}^{n}\frac{(-1)^n\cos(\ln(k))}{k^x}$$ $$B_{n}(x)=\sum_{k=1}^{n}\frac{(-1)^n\sin(\ln(k))}{k^x}$$
Find asymptotic behaviour of the formula: $$C_{n}=\sqrt{A_{n}^{2}(x)+B_{n}^{2}(x)}-(A_{n}(x)\cos(\ln(n+1))+B_{n}(x)\sin(\ln(n+1)))$$ Where $x=const.$ and $x\in(0,1)$
The main purpose of this question is following:
Is this true that $$\frac{1}{(n+2)^{2x}}>\frac{2}{(n+1)^x}C_{n}$$ for $n$ big enough ?
I know that if there is small enough function asymptotically equal to $C_{n}$ then answer for the second question is positive.
Additionaly i got a interesting way to solve a problems like this. The link is below:
However i am not able to apply that in this specific problem.
I will be grateful for answers or even hints.
EDIT: Let $$D_{n}(x)=A_{n}^{2}(x)+B_{n}^{2}(x)$$.
Then the second question is equivalent to the following:
$$\frac{1}{(n+2)^{2x}}>D_{n+1}(x)-(\sqrt{D_{n}(x)}-\frac{1}{(n+1)^{x}})^{2}$$
Is there anybody out there who is able to show me nontrivial function asymptotically equal to right side of the inequality above?
Your previous post indicates you meant $A_{n}(x)=\sum_{k=1}^{n}\frac{(-1)^k\cos(\ln(k))}{k^x}$. Then
$$\sum_{k=1}^{2n-1} (-1)^k k^{-s}= -\eta(s) -\sum_{k=2n}^\infty (-1)^k k^{-s} \\=-\eta(s) -\sum_{k=n}^\infty (2k)^{-s}-(2k+1)^{-s}=-\eta(s)- \sum_{k=n}^\infty \int_{2k}^{2k+1} s x^{-s-1}dx \\= -\eta(s)-\sum_{k=n}^\infty \frac12 \int_{2k}^{2k+2} s x^{-s-1}dx + \frac12\int_{2k}^{2k+1} s (x^{-s-1}-(x+1)^{-s-1})dx$$
$$ = -\eta(s) - \frac12\int_{2n}^\infty s x^{-s-1}dx + \sum_{k=n}^\infty O( \frac12 s (s+1) (2k)^{-s-2}) \\ =-\eta(s) - 2^{-s-1}n^{-s} + O(2^{-s-1} s(s+1) \frac{|n^{-s-1}|}{\Re(s)+1})$$
Take the real and imaginary part with $s = x-i$ to get your $A_n(x),B_n(x)$. For $2n $ instead of $2n-1$ it works the same way up to a sign obtaining
$$\sum_{k=1}^{2n} (-1)^k k^{-s}=-\eta(s) + 2^{-s-1}n^{-s} + O(2^{-s-1} s(s+1) \frac{|n^{-s-1}|}{\Re(s)+1})$$