Convergence of series $ \frac1{n}\sum\limits_{k=1}^n a_k$ Let $ (a_n)$ and $(b_n)$ two real sequences. Further we know that $ b_n \to b$ and 
$$ \lim_n\frac{1}{n}\sum_{k=1}^n (a_k-b_k) \to 0$$
Is it true, and if so, how could I prove?
$$ \lim_n\frac{1}{n}\sum_{k=1}^n a_k \to b$$
Thanks for your help,
 A: First show that $\lim\limits_{n\rightarrow\infty} {1\over n} \sum\limits_{k=1}^n b_k=b$.
Towards this end, let $\epsilon>0$. 
Choose $M$ so that 
$|b_k-b|<\epsilon $ for all $k\ge  M$. Now write
$$
\tag{1}{1\over n} \sum\limits_{k=1}^n\, b_k = {1\over n} \sum\limits_{k=1}^M b_k+{1\over n} \sum\limits_{k=M+1}^n b_k.
$$
Now$$\tag{2}\lim_{n\rightarrow\infty} {1\over n} \sum\limits_{k=1}^M b_k=0.$$ 
Also:
$$ (b-\epsilon){ n- M \over n}\le  {1\over n} \sum\limits_{k=M+1}^n b_k 
\le (b+\epsilon) {n- M  \over n}.$$ 
Taking limits as $n\rightarrow\infty$ of the above gives
$$\tag{3} 
(b-\epsilon) \le \liminf_n\, {1\over n} \sum\limits_{k=M+1}^n b_k 
\quad\text{ and }\quad
\limsup_n\, {1\over n} \sum\limits_{k=M+1}^n b_k 
\le (b+\epsilon) .$$ 
Since $\epsilon$ was an arbitrary positive number,  it follows from (1), (2), and (3) that $\lim\limits_{n\rightarrow\infty}{1\over n} \sum\limits_{k=1}^n b_k=b$.
Now write
$$\eqalign{
\lim_{n\rightarrow\infty} {1\over n} \sum_{k=1}^n a_k
&= \lim_{n\rightarrow\infty}{1\over n} \sum_{k=1}^n (a_k-b_k+b_k)\cr
&=\lim_{n\rightarrow\infty}{1\over n}\sum_{k=1}^n(a_k-b_k)
  + \lim_{n\rightarrow\infty} {1\over n} \sum_{k=1}^n b_k \cr
&=0+b\cr
&=b.
}
$$
