# If $\langle{a_n}\rangle$ is convergent with limit l, how would I show that $\langle{r_n}\rangle$ also converges?

Let $\langle{a_n}\rangle$ be a sequence of real numbers and define the seqence $\langle{r_n}\rangle$ by $$r_n = \frac{1}{n} (a_1 + a_2 + ... + a_n)$$

If $\langle{a_n}\rangle$ is convergent with limit l, how would I show that $\langle{r_n}\rangle$ also converges?

Most of the questions in the exercise I have been doing all relate to one of the following; The comparison test, the ratio test, the root test or the alternating series test. Based on my note taking I have been looking more into the root and comparison tests to see how this sequence converges however I feel like I'm not getting anywhere.

edit: I have also just started learning about absolute and conditional convergence. I have just started working through some theorems as it seems very relatable.

• $\{r_n\}$ is known as the Cesaro means of $\{a_n\}$. The convergence of Cesaro means given the convergence of the original sequence is a well-known result, and it has also been asked here before, e.g. here: math.stackexchange.com/questions/565288/… Nov 20, 2016 at 18:45

The sequence $r_n$ is really a sequence of "averages". So we prove the following statement: If $s_n \to s$, then $\sigma_n \to s$, where $$\sigma_n=\dfrac{s_0+s_1+\cdots+s_n}{n+1}$$ Note that the only difference here is the $n+1$ and that comes from the fact that I begin my sequence at $0$ rather than $1$. Now we prove this statement:

First, observe that

\begin{split} |\sigma_n-s|&=\left| \frac{s_0+s_1+s_2+\cdots+s_n}{n+1}-s\right| \\ &=\left|\frac{s_0+s_1+\cdots+s_n}{n+1}-\frac{(n+1)s}{n+1}\right| \\ &=\left|\frac{(s_0-s)+(s_1-s)+\cdots+(s_n-s)}{n+1} \right| \\ &=\left| \frac{1}{n+1} \sum_{k=0}^n s_k -s\right| \\ &\leq \frac{1}{n+1} \sum_{k=0}^n |s_n-s|= \frac{1}{n+1} \sum_{k=0}^{N-1} |s_n-s| + \frac{1}{n+1} \sum_{k=N}^n |s_n-s| \end{split}

Since $\lim_{n\to \infty} s_n=s$, given $\epsilon>0$ there exists a $N \in \mathbb{N}$ such that $|s_n-s|<\frac{\epsilon}{2}$ for $n\geq N$. Moreover, since $s_n$ is a convergent sequence, it is bounded. Then there exists a $M \in \mathbb{N}$ such that $|s_n-s|\leq M$ for all $n$. Then $$\frac{1}{n+1} \sum_{k=0}^{N-1} |s_n-s|\leq \frac{NM}{n+1}$$ Choose $P \in \mathbb{N}$ such that $P>\frac{2NM}{\epsilon}$. Then this implies $P+1>\frac{2NM}{\epsilon}+1>\frac{2NM}{\epsilon}$. But then $\frac{NM}{P+1}<\frac{\epsilon}{2}$. Therefore, $$\frac{1}{n+1} \sum_{k=0}^{N-1} |s_n-s| < \frac{\epsilon}{2}$$ for $n>P$.

Let $\mathcal{J}=\max\{P,N\}$, then $$|\sigma_n-s|\leq \frac{1}{n+1} \sum_{k=1}^{N-1} |s_n-s| + \frac{1}{n+1} \sum_{k=N}^n |s_n-s|<\frac{\epsilon}{2}+\frac{(n-N+1)\frac{\epsilon}{2}}{n+1}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ for $n>\mathcal{J}$. Therefore, $\sigma_n$ converges to $s$.

Since in your case $s_n \to 1$, $\sigma_n \to 1$. But this idea works more generally as you can see.

Hint. If $a_n\to L$ then for any $\epsilon>0$ there is $N$ such that for $n>N$, $L-\epsilon\leq a_n\leq L+\epsilon$.

Hence for $n>N$, $$\frac{1}{n} \sum_{i=1}^N a_i+ (L-\epsilon)\leq\frac{1}{n} (\sum_{i=1}^N a_i+\sum_{i=N+1}^n a_i)\leq \frac{1}{n} \sum_{i=1}^N a_i+ (L+\epsilon).$$ What happens to both sides when $n$ goes to infinity?

Notice that exists such $m$ that $n > m \Rightarrow |a_n - 1| < \epsilon$, so you can say: $$\sum_{n=m}^{m+k-1} a_n \leq k(1 + \epsilon)$$

Arithmetic mean of this elements is not bigger than $1+\epsilon$.

Can you handle it from here?

Hint: Squeeze theorem.

Hint

let $\epsilon>0$. for $n>0$, we have

$$|r_n-l|\leq\frac{\sum_{k=1}^n|a_k-l|}{n}.$$

but $$\exists N_1>0\:\: \forall k>N_1\; |a_k-l|<\frac{\epsilon}{2}$$

thus, for $n>N_1$,

$$|r_n-l|\leq \frac{\sum_{k=1}^{N_1}|r_k-l|}{n}+\frac{n-N_1}{n} \frac{ \epsilon }{2}$$

$$\implies |r_n-l|\leq V_n+\frac{\epsilon}{2}$$

with $\lim_{n\to+\infty}V_n=0.$

$$\implies \exists N_2>0\;\;:\forall n>N_2\;\; V_n<\frac{\epsilon}{2}$$

finally

$$\forall n>\max(N_1,N_2)\;\; |r_n-l|<\epsilon$$

qed.

Using $\textbf{Stolz Lemma}$:

$$\lim_{n\rightarrow\infty}\frac{a_1+a_2+...+a_n}{n}=\lim_{n\rightarrow\infty}\frac{a_n}{1}=L$$