Объясни подробнее этот шаг и почему он верен в решении ниже
[
\left| \dfrac{1}{n} \sum_{k=N}^n (a_k - a) \right| \leq \dfrac{n - N + 1}{n} \cdot \varepsilon \leq \varepsilon.
]
Пиши на русском, пожалуйста
Решение:
Answer:

No, it is not possible to arrange all the rational numbers of the interval ([0, 1]) into a convergent sequence.

Explanation:

To understand why, let's recall the definition of a convergent sequence and consider the nature of the rational numbers in the interval ([0, 1]).

1. Definition of a Convergent Sequence:

A sequence ({a_n}) is said to converge to a real number (L) if for every (\varepsilon > 0), there exists a natural number (N) such that for all (n \geq N), the terms of the sequence satisfy:

[
|a_n - L| < \varepsilon
]

This means that beyond a certain index (N), all terms of the sequence are arbitrarily close to the limit (L).

2. The Rational Numbers in ([0, 1]):

- The set of rational numbers (\mathbb{Q} \cap [0, 1]) is dense in ([0, 1]). This means that between any two real numbers in ([0, 1]), there are infinitely many rational numbers.
- The set (\mathbb{Q} \cap [0, 1]) is also countable, so we can list its elements in a sequence ({q_n}).

3. Attempting to Arrange All Rationals into a Convergent Sequence:

Suppose, for contradiction, that it is possible to arrange all the rational numbers in ([0, 1]) into a sequence ({q_n}) that converges to some real number (L).

Possible Scenarios:

- Case 1: (L \in [0, 1])
- Case 2: (L < 0) or (L > 1)

We'll consider both cases.

Case 1: (L \in [0, 1])

Since the sequence ({q_n}) contains all rational numbers in ([0, 1]), it must include rationals that are arbitrarily far from (L), because:

- For any (\delta > 0), there exist rationals in ([0, 1]) that are at least (\delta) away from (L).
- This is due to the density of the rationals in ([0, 1]). No matter what (L) is chosen, there are rationals both less than (L - \varepsilon) and greater than (L + \varepsilon) for any (\varepsilon > 0).

Therefore, no matter how large (N) is, there will always be terms (q_n) with (n > N) such that (|q_n - L| \geq \varepsilon). This contradicts the definition of convergence.

Case 2: (L < 0) or (L > 1)

This case is even more straightforward because all terms (q_n) are in ([0, 1]), so:

- (|q_n - L| \geq |L - 1|) if (L > 1)
- (|q_n - L| \geq |L|) if (L < 0)

These distances are positive constants, so the sequence cannot converge to (L) outside ([0, 1]).

4. Conclusion:

In both cases, we arrive at a contradiction. The key reason is:

- A convergent sequence must eventually have all its terms arbitrarily close to its limit.
- Including all rational numbers in ([0, 1]) means the sequence must contain terms that are not arbitrarily close to any single real number.

Therefore, it is impossible to arrange all the rational numbers in ([0, 1]) into a convergent sequence.

Summary:

While the rational numbers in ([0, 1]) are countable and can be listed in a sequence, they cannot be arranged into a sequence that converges. This is because their distribution throughout the interval prevents them from clustering around a single limit point, which is necessary for convergence.


Пусть limn->inf an = a, докажите, что limn->inf (a1 + ... + an)/n = a. Верно ли обратное?


Proof that if (\lim_{n \to \infty} a_n = a), then (\lim_{n \to \infty} \dfrac{a_1 + a_2 + \dots + a_n}{n} = a):

---

1. Understanding the Problem:

We are given a sequence ({a_n}) that converges to (a), i.e.,

[
\lim_{n \to \infty} a_n = a.
]

Our goal is to show that the sequence of arithmetic means of the first (n) terms also converges to (a):

[
\lim_{n \to \infty} A_n = a, \quad \text{where} \quad A_n = \dfrac{a_1 + a_2 + \dots + a_n}{n}.
]

---

2. Proof:

Let’s fix an arbitrary (\varepsilon > 0). Since (\lim_{n \to \infty} a_n = a), by the definition of the limit of a sequence, there exists a natural number (N) such that for all (n \geq N),

[
|a_n - a| < \varepsilon.
]

Now, consider (A_n) for (n \geq N). We can split the sum (S_n = a_1 + a_2 + \dots + a_n) into two parts:

[
S_n = (a_1 + a_2 + \dots + a_{N-1}) + (a_N + a_{N+1} + \dots + a_n).
]

Let’s denote:

- (S_{N-1} = a_1 + a_2 + \dots + a_{N-1}) (the sum of the first (N-1) terms),
- (S_n' = a_N + a_{N+1} + \dots + a_n) (the sum of terms from (N) to (n)).

Thus,

[
A_n = \dfrac{S_n}{n} = \dfrac{S_{N-1} + S_n'}{n} = \dfrac{S_{N-1}}{n} + \dfrac{S_n'}{n}.
]

Estimate the First Term (\dfrac{S_{N-1}}{n}):

Since (S_{N-1}) is a fixed finite sum (constant), and (n \to \infty), we have:

[
\left| \dfrac{S_{N-1}}{n} \right| \leq \dfrac{|S_{N-1}|}{n} \to 0 \quad \text{as} \quad n \to \infty.
]

Estimate the Second Term (\dfrac{S_n'}{n}):

We can write:

[
S_n' = \sum_{k=N}^n a_k = \sum_{k=N}^n [a + (a_k - a)] = (n - N + 1)a + \sum_{k=N}^n (a_k - a).
]

Thus:

[
\dfrac{S_n'}{n} = \dfrac{(n - N + 1)a}{n} + \dfrac{1}{n} \sum_{k=N}^n (a_k - a).
]

Simplify and Estimate:

1. (\dfrac{(n - N + 1)a}{n} = a \left(1 - \dfrac{N - 1}{n} \right) \to a) as (n \to \infty).

2. For the sum involving ((a_k - a)), since (|a_k - a| < \varepsilon) for (k \geq N):

[
\left| \dfrac{1}{n} \sum_{k=N}^n (a_k - a) \right| \leq \dfrac{n - N + 1}{n} \cdot \varepsilon \leq \varepsilon.
]

As (n \to \infty), (\dfrac{n - N + 1}{n} \to 1), so:

[
\left| \dfrac{1}{n} \sum_{k=N}^n (a_k - a) \right| \leq \varepsilon.
]

Combine the Estimates:

Now, combine the estimates for (A_n):

[
\left| A_n - a \right| = \left| \dfrac{S_{N-1}}{n} + \dfrac{(n - N + 1)a}{n} + \dfrac{1}{n} \sum_{k=N}^n (a_k - a) - a \right|.
]

Simplify:

[
\left| A_n - a \right| \leq \left| \dfrac{S_{N-1}}{n} \right| + \left| a \left(1 - \dfrac{N - 1}{n} \right) - a \right| + \left| \dfrac{1}{n} \sum_{k=N}^n (a_k - a) \right|.
]

[
\left| A_n - a \right| \leq \left| \dfrac{S_{N-1}}{n} \right| + \left| a \left( - \dfrac{N - 1}{n} \right) \right| + \varepsilon.
]

As (n \to \infty):

- (\left| \dfrac{S_{N-1}}{n} \right| \to 0),
- (\left| a \left( - \dfrac{N - 1}{n} \right) \right| \to 0),
- (\varepsilon) is arbitrary.

Therefore:

[
\limsup_{n \to \infty} |A_n - a| \leq \varepsilon.
]

Since (\varepsilon > 0) was arbitrary, we conclude that:

[
\lim_{n \to \infty} A_n = a.
]

---

Answer to the Second Part: Is the Converse True?

Conclusion: The converse is not necessarily true.

---

Explanation with Counterexample:

We will construct a sequence ({a_n}) such that the sequence of arithmetic means (A_n) converges to (a), but ({a_n}) itself does not converge to (a).

Example Sequence:

Let’s define the sequence ({a_n}) as:

[
a_n = \begin{cases}
a + 1, & \text{if } n \text{ is even}, \
a - 1, & \text{if } n \text{ is odd}.
\end{cases}
]

Analysis:

1. Behavior of ({a_n}):

The sequence alternates between (a + 1) and (a - 1):

- (a_1 = a - 1),
- (a_2 = a + 1),
- (a_3 = a - 1),
- (a_4 = a + 1), and so on.

Therefore, ({a_n}) does not converge, because it does not approach any single value as (n \to \infty); it keeps oscillating.

2. Behavior of (A_n = \dfrac{a_1 + a_2 + \dots + a_n}{n}):

We will show that (A_n) converges to (a).

- Sum of the First (n) Terms:

For (n) even:

- There are (n/2) odd positions with (a - 1),
- (n/2) even positions with (a + 1).

Sum:

[
S_n = \left( \dfrac{n}{2} \right)(a - 1) + \left( \dfrac{n}{2} \right)(a + 1) = n a.
]

So,

[
A_n = \dfrac{S_n}{n} = \dfrac{n a}{n} = a.
]

For (n) odd:

- There are (\dfrac{n + 1}{2}) odd positions,
- (\dfrac{n - 1}{2}) even positions.

Sum:

[
S_n = \left( \dfrac{n + 1}{2} \right)(a - 1) + \left( \dfrac{n - 1}{2} \right)(a + 1) = n a - 1.
]

So,

[
A_n = \dfrac{S_n}{n} = \dfrac{n a - 1}{n} = a - \dfrac{1}{n} \to a \quad \text{as} \quad n \to \infty.
]

In both cases, (A_n \to a) as (n \to \infty).

Conclusion from Counterexample:

- The sequence of arithmetic means (A_n) converges to (a).
- The original sequence ({a_n}) does not converge.

Therefore, the converse statement is false:

- It is not necessarily true that if (\lim_{n \to \infty} A_n = a), then (\lim_{n \to \infty} a_n = a).

---

Summary:

- Direct Statement ((\lim\limits_{n \to \infty} a_n = a) implies (\lim\limits_{n \to \infty} A_n = a)): True. We proved that if the sequence ({a_n}) converges to (a), then the sequence of its arithmetic means also converges to (a).
- Converse Statement ((\lim\limits_{n \to \infty} A_n = a) implies (\lim\limits_{n \to \infty} a_n = a)): False. We provided a counterexample where ({a_n}) does not converge, but its arithmetic means do converge to (a).

---

Answer:

Yes, if limₙ→∞ aₙ = a, then limₙ→∞ (a₁ + ⋯ + aₙ)⁄n = a; however, the converse is not necessarily true—a sequence’s means may converge to a even if the sequence itself does not.