1

I think I've managed to prove the following theorem, and would like feedback on the proof sketch (correct or not, not rigorous enough etc.)

  • Theorem: $a_n\in \mathbb{R}$ is a sequence, $n\in \mathbb{N}$. If $\lim _{n\to \infty }\left(a_n\right)\:=\:L$, then $\lim _{n\to \infty }\left(\frac{a_1+a_2+...+a_n}{n}\right)=L$

Proof (using the epsilon definition): $$\left|\frac{a_1+a_2+...+a_n}{n}-L\right|\:=\:\left|\frac{a_1+a_2+...+a_n\:-\:n\cdot L}{n}\right|\:=\:\left|\frac{a_1-L+a_2\:-L+...+a_n\:-L}{n}\right|$$

Now, since $\lim _{n\to \infty }\left(a_n\right)\:=\:L$, it means that for every $ɛ>0$ there exists a real number $N$ such as that for every $n>N$, the following is true: $\left|a_n-L\right|<ɛ$.

Picking $N=0$, it follows that: $\left|a_1-L\right|<ɛ, \left|a_2-L\right|<ɛ, \left|a_3-L\right|<ɛ$ and so forth. So: $$\left|\frac{a_1-L+a_2\:-L+...+a_n\:-L}{n}\right|\:<\:\left|\frac{ɛ\:+\:ɛ\:+\:ɛ\:+\:...\:ɛ\:\left(n\:times\right)}{n}\right|\:=\:\left|\frac{n\cdot ɛ}{n}\right|=\left|ɛ\right|=ɛ$$

And thus:

$\left|\frac{a_1+a_2+...+a_n}{n}-L\right|\:<\:ɛ$

∎

I'm feeling like I made a mistake in one of my assumptions in there, or that I did not phrase myself precisely enough. Would be very grateful for a peer review :)

Blabla
  • 169
  • You need to use $|a+b| \le |a|+|b|$ in the last set of equations. – eyeballfrog Nov 10 '22 at 14:06
  • Assumptions for inequalities comes after $N$, so you have not it, for example, for $a_1$. And you cannot pick up $N$. It exists after choosing $\varepsilon$, so, $N = N(\varepsilon)$. – zkutch Nov 10 '22 at 14:08
  • Perhaps the following idea will help: for members of the sequence with a index less than $N$ use boundedness of sequence, which comes from limes existence. – zkutch Nov 10 '22 at 14:25
  • Stolz–Cesàro theorem – NN2 Nov 10 '22 at 14:55

1 Answers1

1

as I wrote above in the comments, we need to slightly improve the idea of your proof:

Let's choose $\forall \varepsilon > 0$. Then $\exists N$, such that, $|a_n-L|<\frac{\varepsilon}{2}$ for $n>N$. Also, $\exists M>0$, such that $|a_n| \leqslant M$ and $|a_n-L|\leqslant 2M$ for all $n$. Then, knowing $\frac 1 n \to 0$, we can write, that $\exists n_0=n_0(\varepsilon)>N$, such that $\frac 1 n < \frac{\varepsilon}{4NM}$ forall $n>n_0$. Using all these inequalities we can write

$$\left|\sum\limits_{k=1}^{n}\frac{a_k}{n}-L\right|=\left|\sum\limits_{k=1}^{n}\frac{a_k}{n}-\sum\limits_{k=1}^{n}\frac{L}{n} \right|\leqslant \sum\limits_{k=1}^{n}\frac 1 n |a_k-L| = \\ =\frac 1 n \cdot |a_1-L| +\cdots + \frac 1 n \cdot |a_N-L| + \frac 1 n \cdot |a_{N+1}-L|+\cdots +\frac 1 n \cdot |a_n-L| \leqslant \\ \leqslant N \cdot 2M \cdot \frac{\varepsilon}{4NM}+\frac{\varepsilon}{2}\left(\overbrace{\frac 1 n +\cdots + \frac 1 n}^{\text{from } N+1 \text{ up to }n}\right)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon$$ forall $n>n_0$, where in last braces we have summunds from $N+1$ up to $n$.

Used idea can be extended to get famous Teplits theorem.

zkutch
  • 13,410