For a bounded sequence $a_n$, $\lim\limits_{k\to\infty}(a_1+...+a_{n_k})/n_k=a$ implies $\lim_{n\to\infty}(a_1+...+a_n)/n=a$. I am reading a paper on Markov chains and am trying to prove a lemma left to the reader that I need for the Ergodic Theorem for Markov Chains (though the lemma requires no knowledge of any of this). The statement of the lemma is as follows:
Let $\{a_n\}$ be a bounded sequence and suppose for $\{n_k\}$ a sequence of integers with $\lim\limits_{k\to\infty}\frac{n_k}{n_{k+1}}=1$ we have $\lim\limits_{k\to\infty} \frac{a_1+...+a_{n_k}}{n_k}=a$. Then $\lim\limits_{n\to\infty} \frac{a_1+....+a_n}{n}=a$.
We can choose $K$ large enough to guarantee that $\left|\frac{a_1+...+a_{n_K}}{n_K}-a\right|<\epsilon$ and choose $n$ even larger than $n_K$ so that
$\left| \frac{(a_1+...+a_n)}{n}-a\right|\leq\left|\frac{a_1+...+a_{n_K}}{n_K}-a\right|+\left|\frac{a_{n_{K+1}}+...+a_n}{n}-a\right|$.
The first absolute value is then less than $\epsilon$ and the second is less than or equal to
$\frac{(M+a)(n-n_K)}{n}$.
This is the piece that I am having trouble showing is less than $\epsilon$. I would greatly appreciate some help making the last leap.
 A: Let $C \le a_n \le D$ for all $n$. By using $a_n -C$ if necessary, we assume $a_n \ge 0$ (that is, $C=0$). For each $n\in \mathbb N$, there is $k$ so that $ n_k\le n <n_{k+1}$. Then 
$$\begin{split}
\frac{a_1 +\cdots + a_n}{n} &\le \frac{a_1 + \cdots + a_n}{n_k} \\
&\le \frac{a_1+\cdots +a_{n_k}}{n_k} + \frac{a_{n_k +1} + \cdots + a_n}{n_k} \\
&\le \frac{a_1+\cdots +a_{n_k}}{n_k} + D\frac{n_{k+1} - n_k}{n_k}\\
&=  \frac{a_1+\cdots +a_{n_k}}{n_k} + D\left(\frac{n_{k+1}}{n_k}-1\right)
\end{split}$$
Thus 
$$\limsup_{n\to \infty}\frac{a_1 +\cdots a_n}{n}  \le \limsup_{k\to \infty}\frac{a_1+\cdots a_{n_k}}{n_k} + D\left(\frac{n_{k+1}}{n_k}-1\right) = a.$$
On the other hand, 
$$\begin{split}
\frac{a_1 + \cdots +a_n}{n} &\ge \frac{a_1 + \cdots + a_n}{n_{k+1}}\\
&\ge \frac{a_1 + \cdots + a_{n_k}}{n_{k+1}}\\
&= \frac{a_1 + \cdots + a_{n_k}}{n_k}\frac{n_k}{n_{k+1}}.
\end{split}$$
Thus 
$$\liminf_{n\to \infty} \frac{a_1 + \cdots +a_n}{n} \ge \liminf_{k\to \infty}\frac{a_1 + \cdots + a_{n_k}}{n_k}\frac{n_k}{n_{k+1}} = a.$$
Thus you are done. 
