Welcome to our community

Be a part of something great, join today!

Marissa043's question on another forum regarding a Cesàro limit

  • Thread starter
  • Moderator
  • #1

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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.
 
  • Thread starter
  • Moderator
  • #2

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 $n\to\infty$.