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

• MHB
• evinda
In summary, the conversation discusses the proof that the limit of a sequence of real numbers converges to a given value. The participants use the definition of limit and the concepts of supremum and infimum to show that the limit exists and is equal to the given value. They also mention the necessity of replacing strict inequalities with non-strict inequalities in the proof.
evinda
Gold Member
MHB
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)

evinda said:
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.)

Janssens said:
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)

evinda said:
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 $\le$.

evinda said:
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.)

evinda said:
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.

Janssens said:
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)

## 1. What does it mean to pick the limit of a sequence?

Choosing the limit of a sequence involves selecting a specific number or value that the terms of the sequence approach as the number of terms increases. This limit can be an actual number or it can be infinity or negative infinity.

## 2. Can the limit of a sequence be any number?

No, the limit of a sequence must be a real number or infinity. It cannot be a complex number or undefined.

## 3. How do you know if a sequence has a limit?

A sequence has a limit if its terms get closer and closer to a specific number or value as the number of terms increases. This can be determined by graphing the sequence or using mathematical methods such as the squeeze theorem or the ratio test.

## 4. Can a sequence have more than one limit?

No, a sequence can only have one limit. If a sequence has multiple limits, it is considered divergent and does not have a well-defined limit.

## 5. Why is it important to pick the limit of a sequence?

The limit of a sequence helps us understand the behavior and trends of the sequence as the number of terms increases. It is also a fundamental concept in calculus and other areas of mathematics, and is crucial in determining convergence or divergence of a series.

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