MHB Doesn't it suffice to pick the limit of the sequence?

[evinda]

Hello! (Wave)

Let $(a_n)$ be a sequence of real numbers such that $a_n \to a$ for some $a \in \mathbb{R}$. I want to show that $\frac{a_1+a_2+\dots+a_n}{n} \to a$.

We have the following:

Let $\epsilon>0$.

Since $a_n \to a$, there is some positive integer $N$ such that if $n \geq N$, then $a-\epsilon<a_n<a+\epsilon$.

Let $b_n=\frac{a_1+a_2+\dots+a_n}{n}$, for $n \geq N$.

We have that $b_n=\frac{a_1+a_2+\dots+a_N}{n}+\frac{a_{N+1}+\dots+a_n}{n}$

and since

$\frac{(n-N)(a-\epsilon)}{n}<\frac{a_{N+1}+\dots+a_n}{n}<\frac{(n-N)(a+\epsilon)}{n}$

we have that

$\frac{C}{n}+\frac{(n-N)(a-\epsilon)}{n}<b_n<\frac{C}{n}+\frac{(n-N)(a+\epsilon)}{n}$

where $C=a_1+a_2+\dots+a_N$.

Can we now just let $n \to +\infty$ ?

Then we would get that $\lim_{n \to +\infty} \left( \frac{C}{n}+ \left( 1-\frac{N}{n}\right)(a-\epsilon)\right)< \lim_{n \to +\infty}b_n < \lim_{n \to +\infty} \left( \frac{C}{n}+\left( 1-\frac{N}{n}\right) (a+\epsilon)\right) \Rightarrow a-\epsilon<\lim_{n \to +\infty} b_n< a+\epsilon \Rightarrow \lim_{n \to +\infty} b_n=a$.

Is this right?
Because I found the proof online and there they pick $\lim_{n \to +\infty} \sup{b_n}$ in order to get the desired result. But is this necessary? (Thinking)

 

[S.G. Janssens]

Can we now just let $n \to +\infty$ ?

Then we would get that $\lim_{n \to +\infty} \left( \frac{C}{n}+ \left( 1-\frac{N}{n}\right)(a-\epsilon)\right)< \lim_{n \to +\infty}b_n < \lim_{n \to +\infty} \left( \frac{C}{n}+\left( 1-\frac{N}{n}\right) (a+\epsilon)\right) \Rightarrow a-\epsilon<\lim_{n \to +\infty} b_n< a+\epsilon \Rightarrow \lim_{n \to +\infty} b_n=a$.

Is this right?
Because I found the proof online and there they pick $\lim_{n \to +\infty} \sup{b_n}$ in order to get the desired result. But is this necessary? (Thinking)

No, I don't think this is entirely correct, unfortunately. Namely, at this point you have not proven that the limit exists. It looks like you have only shown that
$$
a - \epsilon \le \liminf_{n \to \infty}{b_n}, \limsup_{n \to \infty}{b_n} \le a + \epsilon,
$$
for the fixed $\epsilon > 0$ that you started with. This is not sufficient to establish the limit itself, but you are very close! (Namely, $\epsilon > 0$ can be chosen arbitrarily small, and what do you know about the limit of a sequence when its $\liminf$ and $\limsup$ coincide?)

(Also, note that if you take limits, strict inequalities become non-strict inequalities.)

 

[evinda]

No, I don't think this is entirely correct, unfortunately. Namely, at this point you have not proven that the limit exists. It looks like you have only shown that
$$
a - \epsilon \le \liminf_{n \to \infty}{b_n}, \limsup_{n \to \infty}{b_n} \le a + \epsilon,
$$
for the fixed $\epsilon > 0$ that you started with. This is not sufficient to establish the limit itself, but you are very close! (Namely, $\epsilon > 0$ can be chosen arbitrarily small, and what do you know about the limit of a sequence when its $\liminf$ and $\limsup$ coincide?)

(Also, note that if you take limits, strict inequalities become non-strict inequalities.)

So we have that $\frac{C}{n}+\frac{(n-N)(a-\epsilon)}{n}<b_n<\frac{C}{n}+\frac{(n-N)(a+\epsilon)}{n}$ for each $n \geq N$. So it also holds for $\sup{b_n}$. So, $\frac{C}{n}+\frac{(n-N)(a-\epsilon)}{n}<\sup{b_n}<\frac{C}{n}+\frac{(n-N)(a+\epsilon)}{n}$. Right?

Do we know that the limit $\lim_{n \to +\infty} \sup{b_n}$ exists?

Do the equalities $\lim_{n \to +\infty} \sup{b_n}=a$ and $\lim_{n \to +\infty} \inf{b_n}=a$ imply that $\lim_{n \to +\infty} b_n$ exists and is equal to $a$ ? (Thinking)

 

[S.G. Janssens]

So we have that $\frac{C}{n}+\frac{(n-N)(a-\epsilon)}{n}<b_n<\frac{C}{n}+\frac{(n-N)(a+\epsilon)}{n}$ for each $n \geq N$. So it also holds for $\sup{b_n}$. So, $\frac{C}{n}+\frac{(n-N)(a-\epsilon)}{n}<\sup{b_n}<\frac{C}{n}+\frac{(n-N)(a+\epsilon)}{n}$. Right?

Yes, provided that in the second equation, you replace the strict inequalities $<$ with non-strict inequalities $\le$.

Do we know that the limit $\lim_{n \to +\infty} \sup{b_n}$ exists?

Recall that by definition
$$
\limsup_{n \to \infty}{b_n} := \lim_{n \to \infty}{\sup_{k \ge n}{b_k}}, \qquad \liminf_{n \to \infty}{b_n} := \lim_{n \to \infty}{\inf_{k \ge n}{b_k}}.
$$
So you are really taking the limits as $n \to \infty$ of the decreasing sequence $(\sup_{k \ge n}{b_k})_n$ and the increasing sequence $(\inf_{k \ge n}{b_k})_n$. Since both sequences are also bounded, these limits exist. (This is a theorem.)

Do the equalities $\lim_{n \to +\infty} \sup{b_n}=a$ and $\lim_{n \to +\infty} \inf{b_n}=a$ imply that $\lim_{n \to +\infty} b_n$ exists and is equal to $a$ ? (Thinking)

Yes. This is another theorem. You may want to look up these two theorems and their proofs in your analysis notes. (You will be able to follow the proofs.) They are used often.

 

[evinda]

Yes, provided that in the second equation, you replace the strict inequalities $<$ with non-strict inequalities $\le$.
Recall that by definition
$$
\limsup_{n \to \infty}{b_n} := \lim_{n \to \infty}{\sup_{k \ge n}{b_k}}, \qquad \liminf_{n \to \infty}{b_n} := \lim_{n \to \infty}{\inf_{k \ge n}{b_k}}.
$$
So you are really taking the limits as $n \to \infty$ of the decreasing sequence $(\sup_{k \ge n}{b_k})_n$ and the increasing sequence $(\inf_{k \ge n}{b_k})_n$. Since both sequences are also bounded, these limits exist. (This is a theorem.)
Yes. This is another theorem. You may want to look up these two theorems and their proofs in your analysis notes. (You will be able to follow the proofs.) They are used often.

Nice... Thank you... (Smirk)