MHB Marissa043's question on another forum regarding a Cesàro limit

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If the limit of a sequence \( a_n \) approaches \( p \), then the limit of the average \( (a_1 + a_2 + \ldots + a_n)/n \) also approaches \( p \). The proof involves transforming \( a_n \) to \( a_n - p \) to simplify the case to \( p = 0 \). By applying the epsilon-delta definition of limits, it is shown that for sufficiently large \( n \), the average can be expressed as a sum of two parts: a finite sum and a tail that approaches zero. The finite sum can be made arbitrarily small, leading to the conclusion that the average approaches zero. Thus, the Cesàro limit is confirmed as \( s_n \to p \) as \( n \to \infty \).
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Here is the question:
Prove that if $\displaystyle\lim_{n\to\infty}a_n = p$ then $\displaystyle\lim_{n\to\infty}(a_1+a_2+\ldots +a_n)/n = p$.
Here is a link to the question:

Prove this limit.

I have posted a link there to this topic so the OP can find my response.
 
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Hallo Marissa043!

Replacing $a_n$ by $a_n - p$ (for all $n$), you reduce the problem to the case where $p=0$. Now you have to do some real analysis, with an epsilon.So let $\varepsilon>0$. Since $\lim_{n\to\infty}a_n = 0$, there exists $N$ such that $|a_n|<\varepsilon$ whenever $n>N$. Let $s_n = (a_1+a_2+\ldots +a_n)/n$. The idea of the proof is to look at $s_n$, where $n$ is much larger than $N$, by splitting the sum up into the terms from $1$ to $N$ and then from $N+1$ to $n$. In fact, $$|s_n| = \frac1n\biggl|\sum_{k=1}^na_k\biggr| \leqslant \frac1n\biggl|\sum_{k=1}^Na_k\biggr| + \frac1n\biggl|\sum_{k=N+1}^na_k\biggr| \leqslant \frac1n\biggl|\sum_{k=1}^Na_k\biggr| + \frac{\varepsilon(n-N)}n.$$ By taking $n$ large enough, we can make $\frac1n\Bigl|\sum_{k=1}^Na_k\Bigr|$ less than $\varepsilon$, so that $|s_n| < \varepsilon + \frac{\varepsilon(n-N)}n < 2\varepsilon$. Hence $s_n\to0$ as $n\to\infty$.
 
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