Speed of convergence of a special sequence. Let $a\in(0,1)$.
Consider a following sequence:
$$b_{n}=(\sum_{k=1}^{n}\frac{(-1)^k\cos(\ln(k))}{k^{a}})^{2}+(\sum_{k=1}^{n}\frac{(-1)^k\sin(\ln(k))}{k^{a}})^{2}$$

How fast does this series converge to its limit $b$?How to estimate $|b_{n}-b|$?

I know that the limit exists and $b=(\sum_{k=1}^{\infty}\frac{(-1)^k\cos(\ln(k))}{k^{a}})^{2}+(\sum_{k=1}^{\infty}\frac{(-1)^k\sin(\ln(k))}{k^{a}})^{2}$
But apart of it i am not able to tell something else.
I hope for your help.

Hint:
Consider the complement of $b_n$ to $b$, i.e.
$$
\eqalign{
  & \bar b_{\,n}  = \left( {\sum\limits_{k = n + 1}^\infty  {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2}
  + \left( {\sum\limits_{k = n + 1}^\infty  {{{\left( { - 1} \right)^{\,k} \sin \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = \left( {\sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)^{\,k + n} \cos \left( {\ln \left( {k + n} \right)} \right)} \over {\left( {k + n} \right)^{\,a} }}} } \right)^{\,2}
  + \left( {\sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)^{\,k + n} \sin \left( {\ln \left( {k + n} \right)} \right)} \over {\left( {k + n} \right)^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = \left( {{{\left( { - 1} \right)^{\,n} } \over {n^{\,a} }}\sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln n + \ln \left( {1 + k/n} \right)} \right)}
 \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2}
  + \left( {{{\left( { - 1} \right)^{\,n} } \over {n^{\,a} }}\sum\limits_{k = 1}^\infty  {{{\left( { - 1} \right)^{\,k} \sin \left( {\ln n + \ln \left( {1 + k/n} \right)} \right)}
 \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty  {\left( { - 1} \right)^{\,k} {{\cos \left( {\ln n} \right)\cos \left( {\ln \left( {1 + k/n} \right)} \right)
 - \sin \left( {\ln n} \right)\sin \left( {\ln \left( {1 + k/n} \right)} \right)} \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2}  +   \cr 
  &  + {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty  {\left( { - 1} \right)^{\,k} {{\sin \left( {\ln n} \right)\cos \left( {\ln \left( {1 + k/n} \right)} \right)
 + \cos \left( {\ln n} \right)\sin \left( {\ln \left( {1 + k/n} \right)} \right)} \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty  {\left( { - 1} \right)^{\,k} {{\cos \left( {\ln n} \right)\left( {1 - {{k^{\,2} } \over {2n^{\,2} }}
 + O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right)
 - \sin \left( {\ln n} \right)\left( {{k \over n} - {{k^{\,2} } \over {2n^{\,2} }} + O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right)}
 \over {\left( {1 + a{k \over n}   + O\left( {{{k^{\,2} } \over {n^{\,2} }}} \right)} \right)}}} } \right)^{\,2}  +   \cr 
  &  + {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty  {\left( { - 1} \right)^{\,k} {{\sin \left( {\ln n} \right)\left( {1 - {{k^{\,2} } \over {2n^{\,2} }}
 + O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right) + \cos \left( {\ln n} \right)\left( {{k \over n}
 - {{k^{\,2} } \over {2n^{\,2} }} + O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right)} \over {\left( {1 + a{k \over n}
 + O\left( {{{k^{\,2} } \over {n^{\,2} }}} \right)} \right)}}} } \right)^{\,2}  =   \cr 
  &  = \quad  \cdots  \cr} 
$$
It looks that from here you can arrive to the answer to your question.
An alternative approach would be
(here, for simplicity, we consider the development of only the cosine component)
$$
\eqalign{
  & c_{\,2n}  = \left( {\sum\limits_{k = 1}^{2n} {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = \left( {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}}
  + \sum\limits_{k = n + 1}^{2n} {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = \left( {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}}  + {{\left( { - 1} \right)^{\,n} } \over {n^{\,a} }}
\sum\limits_{k = 1}^n {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln n + \ln \left( {1 + k/n} \right)} \right)} \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2}  =   \cr 
  &  = \quad  \cdots  \cr} 
$$
