Asymptotic behaviour of the specific series. Let $a$ be real number. There is a sequence.
$$b_{n}=\sum_{k=n+1}^{\infty}\frac{(-1)^k\cos(\ln (k))}{k^{a}}$$

Find asymptotic behaviour of $b_{n}$. It is find function $f$ so that $b_{n}=O(f(n))$

First of all i know that $b_{n}$ tends to $0$. Apart of it i can't tell anything special.
I hope you will give me some hints.
Regards.
 A: Consider a complex sequence ($i$ is the imaginary unit):
$$c_{n}=\sum_{k=n+1}^{\infty}\frac{(-1)^k k^i}{k^{a}}=\sum_{k=n+1}^{\infty}\frac{(-1)^k}{k^{a-i}}$$
Then $b_n=\Re (c_n)$.
Using the definition of Hurwitz zeta function, we have:
$$c_n= \frac{(-1)^{n+1}}{2^{a-i}} \left(\zeta \left(a-i, \frac{n+1}{2} \right)-\zeta \left(a-i, \frac{n+2}{2} \right) \right)$$
From the integral definition for this function (for $a>0$):
$$\zeta \left(a-i, \frac{n+1}{2} \right)=\frac{1}{\Gamma(a-i)} \int_0^\infty \frac{e^{- \frac{n+1}{2} t} t^{a-1-i} dt}{1-e^{-t}}$$
So we have:
$$c_n= \frac{(-1)^{n+1}}{2^{a-i} \Gamma(a-i)} \int_0^\infty \frac{e^{- \frac{n+1}{2} t} (1-e^{-\frac12 t}) t^{a-1-i} dt}{1-e^{-t}} =\frac{(-1)^{n+1}}{2^{a-i} \Gamma(a-i)} \int_0^\infty \frac{e^{- \frac{n}{2} t} t^{a-1-i} dt}{1+e^{\frac12t}}$$
When $n \to \infty$ only small $t$ contribute to the integral, so let's make a change of variable and then we can do some series expansion:
$$t=\frac{2}{n} u$$

$$c_n= \frac{(-1)^{n+1}}{n^{a-i}\Gamma(a-i)} \int_0^\infty \frac{e^{- u} u^{a-1-i} du}{1+e^{u/n}}$$
This expression is exact, and taking the real part, we can also obtain an exact integral expression for $b_n$.

Asymptotic series for $n \to \infty$ can be obtained in a variety of ways. If $a>1$ then the integral converges absolutely, no matter what we do with the denominator.
Obviously, we can expand:
$$1+e^{u/n}=2 + \frac{u}{n}+ \frac{u^2}{2n^2} + O(u^3/n^3)$$
Taking the first term only, we obtain:

$$c_n \asymp \frac{(-1)^{n+1}}{2n^{a-i}}, \qquad n \to \infty$$

Taking the real part here is easy enough. For more terms we'll have to consider more complicated integrals.
Of course, it might be better to work with the original $t$ integral and use Watson's lemma to obtain explicit asymptotic series.
