# 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?

$$\lim_n\frac{1}{n}\sum_{k=1}^n a_k \to b$$

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You do know the result of $\frac{1}{n}\sum_k^n b_k$? If not, that would be the place to start with. (As $n\rightarrow \infty$ of course). –  user20266 Dec 26 '11 at 13:28
Smells like Cesàro... –  Ｊ. Ｍ. Dec 26 '11 at 13:29
@J.M. He's been dead for over a century now. Interesting that you can still smell him now, and from half a globe away too! –  Asaf Karagila Dec 26 '11 at 20:17
Cesàro is a variety of cheese, @AsafKaragila! –  Bruno Joyal Dec 26 '11 at 22:47

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. }

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Slight mistake: in the inequality between (2) and (3), it should be n-M in the numerator, not n-(M+1). –  user18063 Dec 26 '11 at 19:44
@user18063 Yes, thank you. –  David Mitra Dec 26 '11 at 19:49