Real Sequences#

TODO

TODO

Suppose that \(\{a_n\}\) converges to some \(L \in \mathbb{R}\). Then, by definition, for each \(\varepsilon \gt 0\), there exists some integer \(N\) such that

Choose \(\varepsilon = 1\). The actual choice is irrelevant, it will just result in a different bound. Then,

Let's look at the absolute value of \(a_n\):

Using the triangle inequality, we get

For all \(n \ge N\),

and so \(|a_n| \lt 1 + |L|\) for all \(n \ge N\). This means that the modulus of all sequence terms from the \(N\)-th one onwards is less than \(1 + |L|\). Amongst the first \(N-1\) terms of the sequence, choose the one whose modulus \(M\) is greatest. The moduli of the first \(N-1\) terms are thus all less than or equal to \(M\). Essentially, we have

• \(|a_n| \le M\) for every \(n \lt N\);
• \(|a_n| \lt 1 + |L|\) for every \(n \ge N\).

Let \(B = \max\{M, 1 + |L|\}\). Therefore, \(|a_n| \le B\) for every integer \(n\) and so \(a_n\) is bounded.

TODO

TODO

Let \(\varepsilon \gt 0\). Since \((a_n)_{n \in \mathbb{N}}\) and \((b_n)_{n \in \mathbb{N}}\) are convergent, there exist \(N_a, N_b \in \mathbb{N}\) such that

We also assumed that there is an integer \(N\) such that

It follows then

From this we see that

This is the same as

Let \(A = \displaystyle \lim_{n \to \infty} a_n\) and \(B = \displaystyle \lim_{n \to \infty} b_n\).

Proof of (1):

We have to prove that for each \(\varepsilon \gt 0\), there exists some integer \(N\) such that

The case of \(\alpha = \beta = 0\) is trivial, since then \(|\alpha a_n + \beta b_n - (\alpha A + \beta B)|\) is always zero and is thus smaller than all \(\varepsilon \gt 0\). As for the existence of \(N\) - not only does such an integer \(N\) exist, but there are actually infinitely many such integers, since the inequality holds irrespective of \(N\).

If \(\alpha\) and \(\beta\) are not zero, choose some arbitrary \(\varepsilon' \gt 0\). Since \(a_n \to A\) and \(b_n \to B\), there exist integers \(N_A\) and \(N_B\) such that

$$
\begin{aligned}

|a_n - A| \lt \varepsilon' \qquad \forall n \ge N_A \

|b_n - B| \lt \varepsilon' \qquad \forall n \ge N_B

\end{aligned}
$$

Let \(N = \max \{N_A, N_B\}\). Then both inequalities hold for every \(n \ge N\). Now, we look at

By the triangle inequality, we obtain

For every \(n \ge N\), we know that \(|a_n - A| \lt \varepsilon'\) and \(|b_n - B| \lt \varepsilon'\) and so we get

Since \(|\alpha (a_n - A) + \beta (b_n - B)| \le |\alpha| |a_n - A| + |\beta| |b_n - B|\) and \(|\alpha a_n + \beta b_n - (\alpha A + \beta B)| = |\alpha (a_n - A) + \beta (b_n - B)|\), we get

So far, the argument does not depend on the choice of \(\varepsilon'\) which means that it must be true for all \(\varepsilon' \gt 0\). Thus, we have proven that for each \(\varepsilon = |\alpha| \varepsilon' + |\beta| \varepsilon'\), there exists some integer, namely \(N\), such that

which is what we set out to prove.

Proof of (2):

We have to prove that for each \(\varepsilon \gt 0\), there exists some integer \(N\) such that

Choose some arbitrary \(\varepsilon' \gt 0\). Since \(a_n \to A\) and \(b_n \to B\), there exist integers \(N_A\) and \(N_B\) such that

$$
\begin{aligned}

&|a_n - A| \lt \varepsilon' \qquad \forall n \ge N_A \

&|b_n - B| \lt \varepsilon' \qquad \forall n \ge N_B.

\end{aligned}
$$

Let \(N = \max \{N_A, N_B\}\). Then both inequalities hold for all \(n \ge N\). We transform the expression \(|a_n \cdot b_n - A \cdot B|\) a bit:

Using the triangle inequality, we obtain

which is equivalent to

For all \(n \ge N\),

which in turn means that, for all \(n \ge N\),

Recall that every convergent sequence is bounded. This means that there exists some \(B \gt 0\) and some integer \(\mathcal{N}\) such that \(|b_n| \le B\) for all \(n \ge \mathcal{N}\). Therefore,

Set \(\varepsilon = B \cdot \varepsilon' + |A| \cdot \varepsilon'\). Therefore,

It is obvious that \(\varepsilon\) can be any positive real number. Moreover, since the argument so far does not depend on any particular choice of \(\varepsilon\), we know the argument holds for all \(\varepsilon \gt 0\). We have thus proven that for each \(\varepsilon \gt 0\), there exists some integer, namely \(\mathcal{N}\), such that

which is what we set out to show.

Proof of (3):

TODO

Proof of (4):

TODO

TODO

The following proof was generated by AI and may contain mistakes. TODO: Review

We want to prove that the sequence \(a_n = \frac{1}{n^k}\) converges to \(0\) for any fixed natural number \(k > 0\).
By the definition of convergence, we need to show that for every \(\epsilon > 0\), there exists a natural number \(N\) such that for all \(n > N\), we have \(|a_n - 0| < \epsilon\).

Let \(\epsilon > 0\) be an arbitrary positive real number. We want to find an \(N \in \mathbb{N}\) such that for all \(n > N\),

Since \(n\) is a natural number, \(n \ge 1\). Since \(k\) is a natural number \(k > 0\), we have \(n^k \ge 1^k = 1\). Thus, \(n^k\) is always positive.
Therefore, the absolute value simplifies to:

So, we need to find \(N \in \mathbb{N}\) such that for all \(n > N\),

Since \(\epsilon > 0\) and \(n^k > 0\), we can rearrange the inequality:
Take the reciprocal of both sides (this reverses the inequality sign):

Since \(k > 0\), we can take the \(k\)-th root of both sides. The function \(x \mapsto x^{1/k}\) is strictly increasing for positive \(x\).

We need to find a natural number \(N\) such that for all \(n > N\), the condition \(n > (\frac{1}{\epsilon})^{1/k}\) holds.
By the Archimedean Property of the real numbers, for any positive real number, such as \((\frac{1}{\epsilon})^{1/k}\), there exists a natural number \(N\) greater than it.
So, we can choose \(N\) to be any natural number such that

For example, we can choose \(N = \lfloor (\frac{1}{\epsilon})^{1/k} \rfloor + 1\).

Now, let's verify this choice of \(N\). Suppose \(n\) is any natural number such that \(n > N\). Since \(N > (\frac{1}{\epsilon})^{1/k}\), we have \(n > N > (\frac{1}{\epsilon})^{1/k}\). So, \(n > (\frac{1}{\epsilon})^{1/k}\). Since \(k > 0\), raising both sides to the power of \(k\) (which is an increasing function for positive values) preserves the inequality:

Since \(n^k > 0\) and \(\epsilon > 0\), we can take the reciprocal of both sides, which reverses the inequality:

Since we already established that \(|\frac{1}{n^k} - 0| = \frac{1}{n^k}\), we have shown that for any \(n > N\),

Since we found such an \(N\) for an arbitrary \(\epsilon > 0\), by the definition of convergence, the sequence \(\frac{1}{n^k}\) converges to \(0\).

This completes the proof.

TODO

TODO

TODO

TODO

TODO

We need to prove two things:

• (I) The limit of \((a_n)_{n \in \mathcal{D}}\) is \(L \in \mathbb{R}\), then all of the subsequences of \((a_n)_{n \in \mathcal{D}}\) converge to \(L\).
• (II) If all of the subsequences of \((a_n)_{n \in \mathcal{D}}\) converge to \(L \in \mathbb{R}\), then the limit of \((a_n)_{n \in \mathcal{D}}\) is \(L\).

Proof of (I):

Let \((b_{m})_{m \in f(\mathcal{D})}\) be a subsequence of \((a_n)_{n \in \mathcal{D}}\). By definition, for each \(m\) there exists some \(n \in \mathcal{D}\) such that \(m = f(n)\) and \(b_m = a_m\). Since \((a_n)_{n \in \mathcal{D}}\) converges to \(L\), we know that for each \(\varepsilon \gt 0\) there exists some \(N \in \mathcal{D}\) with

i.e.

Proof of (II):

TODO

TODO

TODO

TODO

TODO

TODO