Convergence of series $ \frac1{n}\sum\limits_{k=1}^n a_k$

Question

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,

Answer 1

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
$$\eqalign{ \lim_{n\rightarrow\infty} {1\over n} \sum_{k=1}^n a_k &= \lim_{n\rightarrow\infty}{1\over n} \sum_{k=1}^n (a_k-b_k+b_k)\cr &=\lim_{n\rightarrow\infty}{1\over n}\sum_{k=1}^n(a_k-b_k) + \lim_{n\rightarrow\infty} {1\over n} \sum_{k=1}^n b_k \cr &=0+b\cr &=b. } $$