We have \(\lim _{i\to \infty } f_i = f\). In other words, the sequence \((f_i)_{i\in \mathbb {N}}\) coefficientwise stabilizes to \(f\). In other words, for each \(n\in \mathbb {N}\),

(by the definition of “coefficientwise stabilizing”).

Now, let \(n\in \mathbb {N}\). Furthermore, let \(k\in \{ 0,1,\ldots ,n\} \). Then, the sequence \(\bigl([x^k] f_i\bigr)_{i\in \mathbb {N}}\) stabilizes to \([x^k] f\) (by 6, applied to \(k\) instead of \(n\)). In other words, there exists some \(N\in \mathbb {N}\) such that

(by the definition of “stabilizing”). Let us denote this \(N\) by \(N_k\). Thus,

Forget that we fixed \(k\). Thus, for each \(k\in \{ 0,1,\ldots ,n\} \), we have defined an integer \(N_k\in \mathbb {N}\) for which 7 holds. Altogether, we have thus defined \(n+1\) integers \(N_0, N_1, \ldots , N_n \in \mathbb {N}\).

Let us set \(P := \max \{ N_0, N_1, \ldots , N_n\} \). Then, of course, \(P\in \mathbb {N}\).

Now, let \(i\geq P\) be an integer. Then, for each \(k\in \{ 0,1,\ldots ,n\} \), we have \(i\geq P = \max \{ N_0, N_1, \ldots , N_n\} \geq N_k\) (since \(k\in \{ 0,1,\ldots ,n\} \)) and therefore \([x^k] f_i = [x^k] f\) (by 7). In other words, each \(k\in \{ 0,1,\ldots ,n\} \) satisfies \([x^k] f_i = [x^k] f\). Renaming the variable \(k\) as \(m\) in this result, we obtain the following:

In other words, \(f_i \overset {x^n}{\equiv } f\) (by the definition of \(x^n\)-equivalence).

Forget that we fixed \(i\). We thus have shown that all integers \(i\geq P\) satisfy \(f_i \overset {x^n}{\equiv } f\). Hence, there exists some integer \(N\in \mathbb {N}\) such that

(namely, \(N = P\)). This proves Lemma 1.462.

This follows directly from Lemma 1.462 applied to the sequence \((f_i)\) and limit \(f\).

This is the exact same statement as Lemma A.85, but applied to the sequence \((g_i)\) and limit \(g\). It follows directly from Lemma 1.462.

By Lemma A.85, there exists \(K\) such that \(f_i \overset {x^n}{\equiv } f\) for all \(i\geq K\). By Lemma A.86, there exists \(L\) such that \(g_i \overset {x^n}{\equiv } g\) for all \(i\geq L\). Set \(P := \max \{ K, L\} \). Then for all \(i\geq P\), both \(i\geq K\) and \(i\geq L\) hold, so both equivalences hold.

Recall that \(\lim _{i\to \infty } f_i = f\). Hence, Lemma 1.462 shows that there exists some integer \(N\in \mathbb {N}\) such that

Let us denote this \(N\) by \(K\). Hence, all integers \(i\geq K\) satisfy \(f_i \overset {x^n}{\equiv } f\).

Thus, we have found an integer \(K\in \mathbb {N}\) such that

Similarly, using \(\lim _{i\to \infty } g_i = g\), we can find an integer \(L\in \mathbb {N}\) such that

Consider these \(K\) and \(L\). Let us furthermore set \(P := \max \{ K, L\} \). Thus, \(P\in \mathbb {N}\).

We shall now show that each integer \(i\geq P\) satisfies \([x^n](f_i g_i) = [x^n](fg)\).

Indeed, let \(i\geq P\) be an integer. Then, \(f_i \overset {x^n}{\equiv } f\) (since \(i\geq P = \max \{ K, L\} \geq K\)), whereas \(g_i \overset {x^n}{\equiv } g\) (since \(i\geq P = \max \{ K, L\} \geq L\)). Hence, we obtain

(by the multiplicativity of \(x^n\)-equivalence, i.e., Theorem 1.328 (b)). In other words,

(by the definition of \(x^n\)-equivalence). Applying this to \(m = n\), we find

Forget that we fixed \(i\). We thus have shown that all integers \(i\geq P\) satisfy \([x^n](f_i g_i) = [x^n](fg)\). Hence, there exists some \(N\in \mathbb {N}\) such that

(namely, \(N = P\)). In other words,

(by the definition of “stabilizes”).

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\), the sequence \(\bigl([x^n](f_i g_i)\bigr)_{i\in \mathbb {N}}\) stabilizes to \([x^n](fg)\). In other words, the sequence \((f_i g_i)_{i\in \mathbb {N}}\) coefficientwise stabilizes to \(fg\) (by the definition of “coefficientwise stabilizing”). In other words, \(\lim _{i\to \infty }(f_i g_i) = fg\).

The same argument—but using the additivity of \(x^n\)-equivalence instead of the multiplicativity—shows that \(\lim _{i\to \infty }(f_i + g_i) = f + g\). Thus, the proof of Proposition 1.463 is complete.

For each \(n\in \mathbb {N}\):

1. By Lemma 1.462, obtain \(K_f\) such that \(f_i \overset {x^n}{\equiv } f\) for \(i\geq K_f\), and \(K_g\) such that \(g_i \overset {x^n}{\equiv } g\) for \(i\geq K_g\).

2. Set \(P := \max \{ K_f, K_g\} \).

3. For \(i\geq P\), by the additivity of \(x^n\)-equivalence, \(f_i + g_i \overset {x^n}{\equiv } f + g\).

4. Extracting the \(n\)-th coefficient: \([x^n](f_i + g_i) = [x^n](f + g)\).

For each \(n\in \mathbb {N}\):

1. By Lemma 1.462, obtain \(K_f\) such that \(f_i \overset {x^n}{\equiv } f\) for \(i\geq K_f\), and \(K_g\) such that \(g_i \overset {x^n}{\equiv } g\) for \(i\geq K_g\).

2. Set \(P := \max \{ K_f, K_g\} \).

3. For \(i\geq P\), by the multiplicativity of \(x^n\)-equivalence, \(f_i \cdot g_i \overset {x^n}{\equiv } f \cdot g\).

4. Extracting the \(n\)-th coefficient: \([x^n](f_i \cdot g_i) = [x^n](f \cdot g)\).

First, let us show that \(g\) is invertible:

Claim 1: The FPS \(g\in K[[x]]\) is invertible.

It remains to prove that \(\lim _{i\to \infty } \frac{f_i}{g_i} = \frac{f}{g}\).

Recall that \(\lim _{i\to \infty } f_i = f\). Hence, Lemma 1.462 shows that there exists some integer \(N\in \mathbb {N}\) such that

Let us denote this \(N\) by \(K\). Hence, all integers \(i\geq K\) satisfy \(f_i \overset {x^n}{\equiv } f\).

Thus, we have found an integer \(K\in \mathbb {N}\) such that

Similarly, using \(\lim _{i\to \infty } g_i = g\), we can find an integer \(L\in \mathbb {N}\) such that

Consider these \(K\) and \(L\). Let us furthermore set \(P := \max \{ K, L\} \). Thus, \(P\in \mathbb {N}\).

We shall now show that each integer \(i\geq P\) satisfies \([x^n] \frac{f_i}{g_i} = [x^n] \frac{f}{g}\).

Indeed, let \(i\geq P\) be an integer. Then, \(f_i \overset {x^n}{\equiv } f\) (since \(i\geq P = \max \{ K, L\} \geq K\)), whereas \(g_i \overset {x^n}{\equiv } g\) (since \(i\geq P = \max \{ K, L\} \geq L\)). Since both FPSs \(g_i\) and \(g\) are invertible (by Claim 1), we thus obtain

(by Theorem 1.328 (e), applied to \(a = f_i\), \(b = f\), \(c = g_i\) and \(d = g\)). In other words,

(by the definition of \(x^n\)-equivalence). Applying this to \(m = n\), we find

Forget that we fixed \(i\). We thus have shown that all integers \(i\geq P\) satisfy \([x^n] \frac{f_i}{g_i} = [x^n] \frac{f}{g}\). Hence, there exists some \(N\in \mathbb {N}\) such that

(namely, \(N = P\)). In other words,

(by the definition of “stabilizes”).

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\), the sequence \(\left([x^n] \frac{f_i}{g_i}\right)_{i\in \mathbb {N}}\) stabilizes to \([x^n] \frac{f}{g}\). In other words, the sequence \(\left(\frac{f_i}{g_i}\right)_{i\in \mathbb {N}}\) coefficientwise stabilizes to \(\frac{f}{g}\) (by the definition of “coefficientwise stabilizing”). In other words, \(\lim _{i\to \infty } \frac{f_i}{g_i} = \frac{f}{g}\). This proves Proposition 1.468 (since we already have shown that \(g\) is invertible).

The family \((f_n)_{n\in \mathbb {N}}\) is summable. In other words, the family \((f_k)_{k\in \mathbb {N}}\) is summable (since this is the same family as \((f_n)_{n\in \mathbb {N}}\)). In other words, for each \(n\in \mathbb {N}\),

(by the definition of “summable”).

Let us define

Let us furthermore set

Now, fix \(n\in \mathbb {N}\). Then, all but finitely many \(k\in \mathbb {N}\) satisfy \([x^n] f_k = 0\) (by 8). In other words, there exists a finite subset \(J\) of \(\mathbb {N}\) such that

Consider this subset \(J\). The set \(J\) is a finite set of nonnegative integers, and thus has an upper bound. In other words, there exists some \(m\in \mathbb {N}\) such that

Consider this \(m\).

Let \(k\in \mathbb {N}\) be such that \(k\geq m+1\). If we had \(k\in J\), then we would have \(k\leq m\) (by 12), which would contradict \(k\geq m+1 {\gt} m\). Thus, we cannot have \(k\in J\). Hence, \(k\in \mathbb {N}\setminus J\) (since \(k\in \mathbb {N}\) but not \(k\in J\)). Therefore, 11 yields \([x^n] f_k = 0\).

Forget that we fixed \(k\). We thus have shown that

Now, let \(i\) be an integer such that \(i\geq m\). Then, \(i\in \mathbb {N}\) (since \(i\geq m\geq 0\)) and \(i+1\geq m+1\) (since \(i\geq m\)). From \(g = \sum _{k\in \mathbb {N}} f_k\), we obtain

On the other hand, from \(g_i = \sum _{k=0}^{i} f_k\), we obtain

Comparing this with 16, we obtain \([x^n] g_i = [x^n] g\).

Forget that we fixed \(i\). We thus have shown that

Hence, there exists some \(N\in \mathbb {N}\) such that

(namely, \(N = m\)). In other words, the sequence \(\bigl([x^n] g_i\bigr)_{i\in \mathbb {N}}\) stabilizes to \([x^n] g\) (by the definition of “stabilizes”).

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\),

In other words, the sequence \((g_i)_{i\in \mathbb {N}}\) coefficientwise stabilizes to \(g\). In other words, \(\lim _{i\to \infty } g_i = g\). In view of 9 and 10, we can rewrite this as

Renaming the summation index \(k\) as \(n\) everywhere in this equality, we can rewrite it as

This proves Theorem 1.475.

The family \((f_n)_{n\in \mathbb {N}}\) is multipliable. In other words, each coefficient in the product of this family is finitely determined (by the definition of “multipliable”).

Let us define

Let us furthermore set

Now, fix \(n\in \mathbb {N}\). Then, the \(x^n\)-coefficient in the product of the family \((f_k)_{k\in \mathbb {N}}\) is finitely determined. In other words, there is a finite subset \(M\) of \(\mathbb {N}\) that determines the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\). Consider this subset \(M\).

The set \(M\) is a finite set of nonnegative integers, and thus has an upper bound. In other words, there exists some \(m\in \mathbb {N}\) such that

Consider this \(m\).

Now, let \(i\) be an integer such that \(i\geq m\). Then, \(i\in \mathbb {N}\) (since \(i\geq m\geq 0\)) and \(m\leq i\). From 19, we see that all \(k\in M\) satisfy \(k\leq m\leq i\) and therefore \(k\in \{ 0,1,\ldots ,i\} \). In other words, \(M\subseteq \{ 0,1,\ldots ,i\} \).

However, the set \(M\) determines the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\). In other words, every finite subset \(J\) of \(\mathbb {N}\) satisfying \(M\subseteq J\subseteq \mathbb {N}\) satisfies

(by the definition of “determines the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\)”). Applying this to \(J = \{ 0,1,\ldots ,i\} \), we obtain

(since \(\{ 0,1,\ldots ,i\} \) is a finite subset of \(\mathbb {N}\) satisfying \(M\subseteq \{ 0,1,\ldots ,i\} \subseteq \mathbb {N}\)).

Moreover, the definition of the product of the multipliable family \((f_k)_{k\in \mathbb {N}}\) shows that

(since \(M\) is a finite subset of \(\mathbb {N}\) that determines the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\)). Comparing this with 20, we find

In view of \(\prod _{k\in \{ 0,1,\ldots ,i\} } f_k = \prod _{k=0}^{i} f_k = g_i\) (by 17) and \(\prod _{k\in \mathbb {N}} f_k = g\) (by 18), we can rewrite this as

Forget that we fixed \(i\). We thus have shown that

Hence, the sequence \(\bigl([x^n] g_i\bigr)_{i\in \mathbb {N}}\) stabilizes to \([x^n] g\).

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\), the sequence \(\bigl([x^n] g_i\bigr)_{i\in \mathbb {N}}\) stabilizes to \([x^n] g\). In other words, the sequence \((g_i)_{i\in \mathbb {N}}\) coefficientwise stabilizes to \(g\). In other words, \(\lim _{i\to \infty } g_i = g\). In view of 17 and 18, we can rewrite this as

Renaming the product index \(k\) as \(n\) everywhere in this equality, we can rewrite it as

This proves Theorem 1.476.

Let a FPS \(a\in K[[x]]\) be written in the form \(a = \sum _{n\in \mathbb {N}} a_n x^n\) with \(a_n\in K\). Then, the family \((a_n x^n)_{n\in \mathbb {N}}\) is summable. Hence, Theorem 1.475 (applied to \(f_n = a_n x^n\)) yields

In other words, \(a = \lim _{i\to \infty } \sum _{n=0}^{i} a_n x^n\). This proves Corollary 1.477.

Let us define

We have assumed that the limit \(\lim _{i\to \infty } \sum _{n=0}^{i} f_n\) exists. Let us denote this limit by \(g\). Thus,

In other words, the sequence \((g_i)_{i\in \mathbb {N}}\) coefficientwise stabilizes to \(g\). In other words, for each \(n\in \mathbb {N}\),

Let \(n\in \mathbb {N}\). Then, the sequence \(\bigl([x^n] g_i\bigr)_{i\in \mathbb {N}}\) stabilizes to \([x^n] g\) (by 23). In other words, there exists some \(N\in \mathbb {N}\) such that

Consider this \(N\). Let \(M\) be the set \(\{ 0,1,\ldots ,N\} \); this is a finite subset of \(\mathbb {N}\).

Let \(i\in \mathbb {N}\setminus M\). Thus,

In other words, \(i\) is a nonnegative integer with \(i\geq N+1\). Hence, \(i-1\geq N\). Therefore, 24 (applied to \(i-1\) instead of \(i\)) yields \([x^n] g_{i-1} = [x^n] g\). But 24 also yields \([x^n] g_i = [x^n] g\) (since \(i\geq N+1\geq N\)). Comparing these two equalities, we find

However, 21 yields

Moreover, 21 (applied to \(i-1\) instead of \(i\)) yields \(g_{i-1} = \sum _{k=0}^{i-1} f_k\). In view of this, we can rewrite 26 as

Hence, \([x^n] g_i = [x^n](f_i + g_{i-1}) = [x^n] f_i + \underbrace{[x^n] g_{i-1}}_{\substack {= [x^n] g_i \\ \text{(by~ \eqref{pf.thm.fps.lim.sum-lim-conv.2})}}} = [x^n] f_i + [x^n] g_i\). Subtracting \([x^n] g_i\) from both sides of this equality, we find \(0 = [x^n] f_i\). In other words, \([x^n] f_i = 0\).

Forget that we fixed \(i\). We thus have shown that all \(i\in \mathbb {N}\setminus M\) satisfy \([x^n] f_i = 0\). Hence, all but finitely many \(i\in \mathbb {N}\) satisfy \([x^n] f_i = 0\) (since all but finitely many \(i\in \mathbb {N}\) belong to \(\mathbb {N}\setminus M\) because the set \(M\) is finite). Renaming the index \(i\) as \(k\) in this statement, we obtain the following: All but finitely many \(k\in \mathbb {N}\) satisfy \([x^n] f_k = 0\).

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\),

In other words, the family \((f_n)_{n\in \mathbb {N}}\) is summable (by the definition of “summable”).

Hence, Theorem 1.475 yields \(\lim _{i\to \infty } \sum _{n=0}^{i} f_n = \sum _{n\in \mathbb {N}} f_n\). In other words, \(\sum _{n\in \mathbb {N}} f_n = \lim _{i\to \infty } \sum _{n=0}^{i} f_n\). This completes the proof of Theorem 1.478.

Let us define

We have assumed that the limit \(\lim _{i\to \infty } \prod _{n=0}^{i} f_n\) exists. Let us denote this limit by \(g\). Thus,

Let \(n\in \mathbb {N}\) be arbitrary. Lemma 1.462 (applied to \(g_i\) and \(g\) instead of \(f_i\) and \(f\)) shows that there exists some integer \(N\in \mathbb {N}\) such that

(since \(g = \lim _{i\to \infty } g_i\)). Consider this \(N\).

Let \(M = \{ 0,1,\ldots ,N\} \). Thus, \(M\) is a finite subset of \(\mathbb {N}\). We shall show that this subset \(M\) determines the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\).

For this purpose, we first observe that \(M = \{ 0,1,\ldots ,N\} \), so that

(since \(g_N\) was defined to be \(\prod _{k=0}^{N} f_k\)). Applying 29 to \(i = N\), we obtain \(g_N \overset {x^n}{\equiv } g\) (since \(N\geq N\)). Therefore, \(g \overset {x^n}{\equiv } g_N\) (since \(\overset {x^n}{\equiv }\) is an equivalence relation and thus symmetric).

Next, we shall show two claims:

Claim 1: For each integer \(j {\gt} N\), we have \(g \overset {x^n}{\equiv } g f_j\).

Claim 2: If \(U\) is any finite subset of \(\mathbb {N}\) satisfying \(M\subseteq U\), then \(g \overset {x^n}{\equiv } \prod _{k\in U} f_k\).

Now, let \(J\) be a finite subset of \(\mathbb {N}\) satisfying \(M\subseteq J\subseteq \mathbb {N}\). Then, \(g \overset {x^n}{\equiv } \prod _{k\in J} f_k\) (by Claim 2, applied to \(U = J\)). In other words,

(by the definition of \(x^n\)-equivalence). Applying this to \(m = n\), we find

The same argument can be applied to \(M\) instead of \(J\) (since \(M\subseteq M\subseteq \mathbb {N}\)), and thus yields

Comparing these two equalities, we find

Forget that we fixed \(J\). We have now shown that every finite subset \(J\) of \(\mathbb {N}\) satisfying \(M\subseteq J\subseteq \mathbb {N}\) satisfies

In other words, the set \(M\) determines the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\). Hence, the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\) is finitely determined (since \(M\) is a finite subset of \(\mathbb {N}\)).

Forget that we fixed \(n\). We thus have shown that for each \(n\in \mathbb {N}\), the \(x^n\)-coefficient in the product of \((f_k)_{k\in \mathbb {N}}\) is finitely determined. In other words, each coefficient in the product of \((f_k)_{k\in \mathbb {N}}\) is finitely determined. In other words, the family \((f_k)_{k\in \mathbb {N}}\) is multipliable. In other words, the family \((f_n)_{n\in \mathbb {N}}\) is multipliable.

Hence, Theorem 1.476 yields \(\lim _{i\to \infty } \prod _{n=0}^{i} f_n = \prod _{n\in \mathbb {N}} f_n\). In other words, \(\prod _{n\in \mathbb {N}} f_n = \lim _{i\to \infty } \prod _{n=0}^{i} f_n\). This completes the proof of Theorem 1.479.

Let \(N_1\) and \(N_2\) be the stabilization indices for \(\ell _1\) and \(\ell _2\) respectively. Set \(N := \max (N_1, N_2)\). Then \(a_N = \ell _1\) and \(a_N = \ell _2\), whence \(\ell _1 = \ell _2\).

This is a direct restatement of Theorem 1.475.

This is a direct restatement of Theorem 1.476.

This is a direct restatement of the summability part of Theorem 1.478.

By Theorem 1.475, the partial sums converge to the infinite sum. By uniqueness of limits (Lemma 1.454), the limit equals the infinite sum.

This is a direct restatement of the multipliability part of Theorem 1.479.

By Theorem 1.476, the partial products converge to the infinite product. By uniqueness of limits (Lemma 1.454), the limit equals the infinite product.