Convergence of series

5.2 Convergence of series

Properties of series. Assuming $\sum a_{n}, \sum b_{n}$ is convergent, then

$\forall λ, μ \in R$ or $C$ , $\sum (λ a_{n} + μ b_{n})$ is convergent, and
$∑ ∑ ∑ (λan + μbn) = λ an + μ bn$
If $a_{n}, b_{n} \in R$ and $a_{n} \leq b_{n}$ , $\forall n \geq 1$ , then $\sum a_{n} \leq \sum b_{n}$ .

Theorem 5.2.1 (Cauchy Criterion) Assuming $(V, ∥ \cdot ∥)$ is complete (any Cauchy sequence in $V$ is convergent), then $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent if and only if ${S_{N}}$ is a Cauchy sequence, i.e. $\forall ϵ > 0$ , $\exists N_{ϵ}$ s.t. $\forall N \geq N_{ϵ}$ , $\forall p \geq 1$ , $∥ S_{N + p} - S_{N} ∥ < ϵ$ , i.e. $∥ a_{N + 1} + \dots + a_{N + p} ∥ < ϵ$ .

Corollary 5.2.2 $(R, | \cdot |)$ and $(C, | \cdot |)$ are both complete, so $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent $\Leftrightarrow$ it satisfies Cauchy condition.

Corollary 5.2.3 If $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent, then $lim_{n \to + \infty} a_{n} = 0$ .

Example 5.2.1 $\sum_{n = 0}^{+ \infty} x^{n}, x \in C$ is divergent when $| x | \geq 1$ , since $| x^{n} | = | x |^{n} \geq 1$ . So $\sum x^{n}$ is convergent if and only if $| x | < 1$ .

Definition 5.2.4 Series $\sum_{n = 1}^{+ \infty} a_{n}$ is absolutely convergent if $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is convergent.

Definition 5.2.5 Series $\sum_{n = 1}^{+ \infty} a_{n}$ is conditionally convergent if it’s convergent yet not absolutely convergent.

Theorem 5.2.6 Absolutely convergent series is convergent.

Proof Proved by Cauchy. Assuming $\sum_{n = 1}^{+ \infty} a_{n}$ is absolutely convergent, $\forall ϵ > 0$ , $\exists N_{ϵ}$ such that $\forall N \geq N_{ϵ}$ , $\forall p \geq 1$ , we have $∥ a_{N + 1} ∥ + \dots + ∥ a_{N + p} ∥ < ϵ$ . So

∥a + ⋅⋅⋅ + a ∥ ≤ ∥a ∥ + ⋅⋅⋅ + ∥a ∥ < 𝜖 N+1 N+p N+1 N+p

meaning $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent. $◻$

Theorem 5.2.7 (Comparison test) If $\exists N_{0}$ s.t. $\forall n \geq N_{0}$ , $∥ a_{n} ∥ \leq ∥ b_{n} ∥$ , then

If $\sum_{n = 1}^{+ \infty} ∥ b_{n} ∥$ is convergent, then $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is convergent.
If $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is divergent, then $\sum_{n = 1}^{+ \infty} ∥ b_{n} ∥$ is divergent.

Proof Proved by Cauchy. Assuming $\sum_{n = 1}^{+ \infty} b_{n}$ is absolutely convergent, $\forall ϵ > 0$ , $\exists N_{ϵ}$ such that $\forall N \geq N_{ϵ}$ , $\forall p \geq 1$ , we have $∥ b_{N + 1} ∥ + \dots + ∥ b_{N + p} ∥ < ϵ$ . So

∥aN+1 ∥ + ⋅⋅⋅ + ∥aN+p ∥ ≤ ∥bN+1∥ + ⋅⋅⋅ + ∥bN+p∥ < 𝜖

meaning $\sum_{n = 1}^{+ \infty} a_{n}$ is absolutely convergent. $◻$

Corollary 5.2.8 If $∥ a_{N} ∥ = O (∥ b_{N} ∥), N \to + \infty$ , if $\sum ∥ b_{n} ∥$ is convergent, then $\sum ∥ a_{n} ∥$ is convergent.

Proof $\exists M_{0}, N_{0}$ such that $\forall n \geq N_{0}$ , $∥ a_{n} ∥ \leq M_{0} ∥ b_{n} ∥$ . Then it’s left as exercise. :) $◻$

Note

Limit.
$lim ∥an∥-< + ∞ ⇒ ∥an∥ = 𝒪(∥bn∥ ),n → + ∞ n→+ ∞ ∥bn∥$
Simultaneous convergence and divergence. If
$∥an-∥ 0 < nl→im+ ∞ ∥bn∥ < + ∞$
, then $\sum a_{n}$ is absolutely convergent $\Leftrightarrow$ $\sum b_{n}$ is absolutely convergent.

Theorem 5.2.9 (D’Alembert) Assuming

lim ∥an+1-∥ = ρ n→+ ∞ ∥an∥

then

If $ρ < 1$ , $\sum a_{n}$ is absolutely convergent.
If $ρ > 1$ , $\sum a_{n}$ is divergent.

Proof

When $ρ < 1$ , take $r \in (ρ, 1)$ , we have
$∥an+1∥ lim -------< r n→+ ∞ ∥an∥$
then $\exists N_{0}$ such that $\forall n \geq N_{0}$ , $\frac{∥ a_{n + 1} ∥}{∥ a_{n} ∥} < r$ . Hence,
$∥a ∥ = -∥an-∥- ⋅ ⋅⋅⋅ ⋅ ∥aN0+1-∥∥a ∥ ≤ ∥a ∥rn−N0 n ∥an −1∥ ∥aN0∥ N0 N0$
meaning $∥ a_{n} ∥ = O (r^{n}), n \to + \infty$ . $\sum_{n = 1}^{+ \infty} r^{n}$ is convergent, so $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is convergent.
When $ρ > 1$ , $\exists N_{0}$ such that $\forall n \geq N_{0}$ , $\frac{∥ a_{n + 1} ∥}{∥ a_{n} ∥} > 1$ , then $∥ a_{n} ∥ \geq ∥ a_{N_{0}} ∥ > 0$ , meaning $\sum a_{n}$ is divergent.

$◻$

Theorem 5.2.10 (Cauchy root) Assuming

∘ ----- lim n ∥an∥ = ρ n→+ ∞

then

If $ρ < 1$ , $\sum a_{n}$ is absolutely convergent.
If $ρ > 1$ , $\sum a_{n}$ is divergent.

Proof

When $ρ < 1$ , take $r \in (ρ, 1)$ , we have
$∘ ----- lim n ∥an∥ < r n→+ ∞$
then $\exists N_{0}$ such that $\forall n \geq N_{0}$ , $\sqrt[n]{∥ a_{n} ∥} < r$ , i.e. $∥ a_{n} ∥ < r^{n}$ . So $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is convergent.
Left as exercise. :)

$◻$

Example 5.2.2 Assuming $a_{n} = \frac{x^{n}}{n!}$ where $x \in C$ or $x$ is a square matrix. Notice that

xn ∥x∥n ∥an ∥ = ∖left∥---∖right∥ ≤ -----= bn n! n!

Since

bn+1 ∥x ∥ n→+ ∞ -----= ------ −−−−→ 0 < 1 bn n + 1

According to D’Alembert, $\sum b_{n}$ is convergent, so $\sum a_{n}$ is absolutely convergent.

Superior limit and inferior limit. Assuming

{a_{n}}

is bounded,

{a_{n}}

has convergent subsequences. The supremum of the limit of all convergent subsequences is called the superior limit, correspondingly the infimum of the limit of all convergent subsequences is called the inferior limit, marked as

\underset{n \to + \infty}{lim sup} a_{n}

and

\underset{n \to + \infty}{lim inf} a_{n}

Generally, we have

----- ----- lim inf∥an+1-∥ ≤ lim inf n∘ ∥a ∥ ≤ lim sup ∘n ∥a ∥ ≤ lim sup ∥an+1∥- n→+ ∞ ∥an ∥ n→+ ∞ n n→+ ∞ n n→+ ∞ ∥an∥

so Cauchy root criterion is stronger than D’Alembert criterion, yet D’Alembert criterion is easier to compute.

Theorem 5.2.11 (Integral criterion) Assuming $f : [1, + \infty) \to R$ is monotonously decreasing, $f (x) > 0$ , then $\sum_{n = 1}^{+ \infty} f (n)$ is convergent $\Leftrightarrow$ $\int_{1}^{+ \infty} f (x) d x$ is convergent.

Example 5.2.3 $\sum \frac{1}{n^{p}}$ where $p > 0$ . At this time both Cauchy root and D’Alembert criterion are both disabled. Yet $\frac{1}{x^{p}}$ monotonously decreases when $p > 0$ , and

∫ +∞ 1 ---dx 1 xp

+∑∞ 1 -p- n=1 n

Similarly, we could seek the convergence and divergence of

∑ 1 ∑ 1 ------p, -------------p,⋅⋅⋅ n(ln n) n lnn (ln ln n)

Example 5.2.4 Seek the convergence and divergence of $\sum_{n = 1}^{+ \infty} [n \ln (\frac{2 n + 1}{2 n - 1}) - 1]$ Notice that $\begin{aligned} a_{n} & = n \ln (\frac{2 n + 1}{2 n - 1}) - 1 = n \ln (1 + \frac{1}{2 n}) - n \ln (1 - \frac{1}{2 n}) - 1 \\ = n [\frac{1}{2 n} - \frac{1}{8 n^{2}} + \frac{1}{24 n^{3}} + \frac{1}{2 n} + \frac{1}{8 n^{2}} + \frac{1}{24 n^{3}} + o (\frac{1}{n^{3}})] - 1 \\ = \frac{1}{12 n^{2}} + o (\frac{1}{n^{2}}) \end{aligned}$

Therefore $a_{n} = O (\frac{1}{n^{2}})$ . Since $\sum_{n = 1}^{+ \infty} \frac{1}{n^{2}}$ is convergent, meaning $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent.

Notice that for $\sum_{n = 1}^{+ \infty} \frac{1}{n^{p}}$ is convergent when $p > 1$ , divergent when $p \leq 1$ . Assuming $a_{n} = \frac{1}{n^{p}}$ , then $\frac{a_{n}}{a_{n + 1}} = \frac{(n + 1)^{p}}{n^{p}} = {(1 + \frac{1}{n})}^{p} = 1 + \frac{p}{n} + o (\frac{1}{n})$ indicating that $p = n (\frac{a_{n}}{a_{n + 1}} - 1) + o (1)$ This introduces Raabe criterion.

Theorem 5.2.12 (Raabe) Assuming $lim_{n \to + \infty} n (\frac{∥ a_{n} ∥}{∥ a_{n + 1} ∥} - 1) = p$ then

1.: When $p > 1$ , $\sum_{n = 1}^{+ \infty} a_{n}$ is absolutely convergent.
2.: When $p < 1$ , $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is divergent.

Proof

1.: Take $r, q$ such that $1 < r < q < p$ , since $lim_{n \to + \infty} n (\frac{∥ a_{n} ∥}{∥ a_{n + 1} ∥} - 1) = p > q$ meaning $\exists N_{1}$ such that $\forall n \geq N_{1}$ , $n (\frac{∥ a_{n} ∥}{∥ a_{n + 1} ∥} - 1) > q$ Since $\frac{∥ a_{n} ∥}{∥ a_{n + 1} ∥} > 1 + \frac{q}{n} > 1 + \frac{r}{n} + o (\frac{1}{n}) = {(1 + \frac{1}{n})}^{r}$ meaning $\exists N_{2} > N_{1}$ such that $\forall n \geq N_{2}$ , $\frac{∥ a_{n} ∥}{∥ a_{n + 1} ∥} > {(1 + \frac{1}{n})}^{r} \Rightarrow n^{r} ∥ a_{n} ∥ > (n + 1)^{r} ∥ a_{n + 1} ∥ \Rightarrow ∥ a_{n} ∥ < \frac{N_{2}^{r} ∥ a_{N_{2}} ∥}{n^{r}}$ So $∥ a_{n} ∥ = O (\frac{1}{n^{r}})$ . Since $\sum_{n = 1}^{+ \infty} \frac{1}{n^{r}}$ is convergent. $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is convergent.
2.: Left as exercise. :)

$◻$

Conditional convergence: $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent, yet $\sum_{n = 1}^{+ \infty} ∥ a_{n} ∥$ is divergent. For alternating series $\sum_{n = 1}^{+ \infty} (- 1)^{n} b_{n}$ where $b_{n} \geq 0$ , we have Leibniz criterion.

Theorem 5.2.13 (Leibniz) Assuming $b_{n} \geq 0$ and is monotonously non-increasing, then we have $\sum_{n = 1}^{+ \infty} (- 1)^{n} b_{n}$ is convergent $\Leftrightarrow b_{n} \to 0$ .

Proof

$◻$

Example 5.2.5 $\sum_{n = 1}^{+ \infty} \frac{(- 1)^{n - 1}}{n^{α}}$ is convergence $\Leftrightarrow α > 0$ . Seek the convergence and divergence of $\sum_{n = 1}^{+ \infty} \frac{(- 1)^{n - 1}}{n^{α} + (- 1)^{n - 1}}$ Notice that $\begin{aligned} a_{n} & = \frac{(- 1)^{n - 1}}{n^{α} + (- 1)^{n - 1}} = \frac{(- 1)^{n - 1}}{n^{α}} [\frac{1}{1 + \frac{(- 1)^{n - 1}}{n^{α}}}] = \frac{(- 1)^{n}}{n^{α}} [1 - \frac{(- 1)^{n - 1}}{n^{α}} + o (\frac{1}{n^{α}})] \\ = \frac{(- 1)^{n - 1}}{n^{α}} \underset{b_{n}}{\underset{⏟}{- \frac{1}{n^{2 α}} + o (\frac{1}{n^{2 α}})}} \end{aligned}$

Since $lim_{n \to + \infty} n^{2 α} b_{n} = 1$ , meaning $\sum_{n = 1}^{+ \infty} \frac{1}{n^{2 α}}$ and $\sum_{n = 1}^{+ \infty} b_{n}$ have the same convergence and divergence. So $\begin{matrix} \frac{(- 1)^{n - 1}}{n^{α}} & b_{n} \\ α \in (0, 0.5] & C.C. & D.V. \\ α \in (0.5, 1] & C.C. & A.C. \\ α \in (1, + \infty) & A.C. & A.C. \end{matrix}$ Terminally, $α \in (0, 0.5]$ , $\sum_{n = 1}^{\infty} a_{n}$ is divergent; $α \in (0, 5, 1]$ , $\sum_{n = 1}^{\infty} a_{n}$ is conditionally convergent; $α \in (1, + \infty)$ , $\sum_{n = 1}^{\infty} a_{n}$ is absolutely convergent.

Dirichlet and Abel criterion.

Theorem 5.2.14 (Dirichlet/Abel) If ${a_{n}} \subseteq V, {b_{n}} \subseteq R$ satisfy one of the following condition,

(Dirichlet) $A_{N} = \sum_{n = 1}^{N} a_{n}$ is bounded, $b_{n}$ monotonously goes to $0$ .
(Abel) $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent, $b_{n}$ is monotonous and bounded.

then $\sum_{n = 1}^{+ \infty} a_{n} b_{n}$ is convergent.

Proof

(Dirichlet) Assuming $b_{n} > 0$ , otherwise consider $- b_{n}$ . Notice that $\begin{aligned} \sum_{k = n + 1}^{n + p} a_{k} b_{k} & = \sum_{k = n + 1}^{n + p} (A_{k} - A_{k - 1}) b_{k} = A_{n + p} b_{n + p} - A_{n} b_{n} - \sum_{k = n + 1}^{n + p} A_{k - 1} (b_{k} - b_{k - 1}) \\ ‖ \sum_{k = n + 1}^{n + p} a_{k} b_{k} ‖ & \leq ∥ A_{n + p} ∥ | b_{n + p} | + ∥ A_{n} ∥ | b_{n} | + \sum_{k = n + 1}^{n + p} ∥ A_{k - 1} ∥ | b_{k} - b_{k - 1} | \\ \leq M b_{n + p} + M b_{n} + \sum_{k = n + 1}^{n + p} M (b_{k - 1} - b_{k}) = 2 M b_{n} \to 0 \end{aligned}$

meaning ${A_{N}}$ is a Cauchy sequence, so $\sum_{n = 1}^{+ \infty} a_{n} b_{n}$ is convergent.
Dirichlet $\Rightarrow$ Abel. Let $c_{n} = b_{n} - lim_{n \to + \infty} b_{n} = b_{n} - b$ , then $c_{n}$ monotonously goes to $0$ . Since $\sum_{n = 1}^{N} a_{n}$ is bounded, according to Dirichlet, $\sum_{n = 1}^{+ \infty} a_{n} c_{n}$ is convergent. Therefore, $\sum_{n = 1}^{+ \infty} a_{n} b_{n} = \sum_{n = 1}^{+ \infty} a_{n} (b + c_{n}) = b \sum_{n = 1}^{\infty} a_{n} + \sum_{n = 1}^{+ \infty} a_{n} c_{n}$ is convergent.

$◻$

Example 5.2.6 Seek the convergence and divergence of $I = \sum_{n = 1}^{+ \infty} \frac{z^{n}}{n}$ where $z \in C$ .

According to Cauchy root method, $\sqrt[n]{\frac{| z^{n} |}{n}} = \frac{| z |}{\sqrt[n]{n}} \overset{n \to + \infty}{\to} | z |$ So when $| z | < 1$ , $I$ is absolutely convergent; when $| z | > 1$ , $I$ is divergent; when $| z | = 1$ , we have

$z = 1$ , $I$ is divergent.
$| z | = 1$ and $z \neq 1$ , let $a_{n} = z_{n}$ where $| \sum_{n = 1}^{N} z^{n} | = | \frac{z - z^{N + 1}}{1 - z} | \leq \frac{2}{| 1 - z |}$ and $b_{n} = \frac{1}{n}$ which monotonously goes to $0$ , according to Dirichlet, $\sum_{n = 1}^{+ \infty} \frac{z^{n}}{n}$ is convergent, and is conditionally convergent.
When $| z | = 1$ , let $z = \cos θ + i \sin θ$ , we have $z^{n} = \cos n θ + i \sin n θ$ , then $\sum_{n = 1}^{+ \infty} \frac{z^{n}}{n} = \sum_{n = 1}^{+ \infty} \frac{\cos n θ}{n} + i \sum_{n = 1}^{+ \infty} \frac{\sin n θ}{n}$ 2 series on R.H.S. are both conditionally convergent when $z \neq \pm 1$ .

Associative law. $\underset{b_{1}}{\underset{⏟}{a_{1}}} + \underset{b_{2}}{\underset{⏟}{a_{2} + a_{3}}} + \underset{b_{3}}{\underset{⏟}{a_{4} + a_{5} + a_{6} + a_{7}}} + \underset{\dots}{\underset{⏟}{\dots}}$ Therefore the partial sum sequence ${B_{N}}$ of ${b_{n}}$ is the subsequence of the partial sum sequence ${A_{N}}$ of ${a_{n}}$ . When the limit of ${A_{N}}$ exists (i.e. $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent), then the limit of ${B_{N}}$ exists (i.e. $\sum_{n = 1}^{+ \infty} b_{n}$ is convergent), and both limits are equivalent. Therefore,

Theorem 5.2.15 Assuming $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent, then the associative law of $\sum_{n = 1}^{+ \infty} a_{n}$ holds.

\sum_{n = 1}^{+ \infty} a_{n}

is divergent, then associative law doesn’t hold, like

\begin{aligned} 1 + \overset{⏞}{(- 1) + 1} + \overset{⏞}{(- 1) + 1} + \dots & = 1 \\ \underset{⏟}{1 + (- 1)} + \underset{⏟}{1 + (- 1)} + 1 + \dots & = 0 \end{aligned}

Commutative law. For conditional convergence, we have

Theorem 5.2.16 (Riemann rearrangement) Assuming $\sum_{n = 1}^{+ \infty} a_{n}, {a_{n}} \subseteq R$ is conditionally convergent, then $\forall A \in R \cup {\pm \infty}$ , $\exists$ a bijection $σ : N^{*} \to N^{*}$ s.t. $\sum_{n = 1}^{+ \infty} a_{n} = A$ .

Proof ${a_{n}}$ has infinite non-negative terms ${b_{n}}$ and negative terms ${c_{n}}$ where $b_{n}, c_{n} \overset{n \to + \infty}{\to} 0$ and $\sum_{n = 1}^{+ \infty} b_{n}, \sum_{n = 1}^{+ \infty} c_{n}$ are both divergent. See Figure 5.2 for hint, then its left as exercise. :) $◻$

Figure 5.2: Proof of Riemann rearrangement

For absolute convergence, we have

Theorem 5.2.17 Assuming $\sum_{n = 1}^{+ \infty} a_{n}$ is absolutely convergent with sum $S$ , then for any arrangement (bijective) $σ : N^{*} \to N^{*}$ , series $\sum_{n = 1}^{+ \infty} a_{σ (n)}$ is also absolutely convergent with sum $S$ .

Proof Mark $S_{n} = \sum_{k = 1}^{n} a_{k}$ and ${\tilde{S}}_{n} = \sum_{k = 1}^{n} a_{σ (k)}$ . Then $\forall ϵ > 0$ , since $\sum_{n = 1}^{+ \infty} a_{n}$ is absolutely convergent, so $\exists n_{ϵ}$ s.t. $\forall n \geq n_{ϵ}$ , $∥ S_{n} - S ∥ = ‖ \sum_{k = n + 1}^{+ \infty} a_{k} ‖ \leq \sum_{k = n + 1}^{+ \infty} ∥ a_{k} ∥ < ϵ$ For any $n$ , mark $A_{n} = {1, 2, . . ., n}$ , $N_{n} = max σ^{- 1} (A_{n}) = max {σ^{- 1} (1), . . ., σ^{- 1} (n)}$ . then ${σ^{- 1} (1), . . ., σ^{- 1} (n)} \subseteq {1, 2, . . ., N_{n}}$ , i.e. $σ_{- 1} (A_{n}) \subseteq A_{N_{n}}$ . Then $\forall m \geq N_{n}$ , we have $σ^{- 1} (A_{n}) \subseteq A_{N_{n}} \subseteq A_{m}$ , i.e. $A_{n} \subseteq σ (A_{m})$ since $σ$ is bijective. Therefore, $\begin{aligned} {\tilde{S}}_{m} & = \sum_{k = 1}^{m} a_{σ (k)} = \sum_{j \in σ (A_{m})} a_{j} + \sum_{j \in σ (A_{m}) ∖ A_{n}} a_{j} = S_{n} + \sum_{j \in σ (A_{m}) ∖ A_{n}} a_{j} \\ ∥ {\tilde{S}}_{m} - S ∥ & \leq ∥ {\tilde{S}}_{m} - S ∥ + ∥ S_{n} - S ∥ \leq ‖ \sum_{j \in σ (A_{m}) ∖ A_{n}} a_{j} ‖ + ϵ \leq \sum_{k = n + 1}^{+ \infty} ∥ a_{j} ∥ + ϵ < 2 ϵ \end{aligned}$

Terminally, $\forall m \geq N_{n_{ϵ}}$ , $∥ {\tilde{S}}_{m} - S ∥ < 2 ϵ$ , meaning $\sum_{n = 1}^{+ \infty} a_{σ (k)}$ is convergent with sum $S$ .

Repeat the proof above with $\sum_{k = 1}^{+ \infty} ∥ a_{σ (k)} ∥$ , we could prove that $\sum_{k = 1}^{+ \infty}$ is absolutely convergent. $◻$

The product of series. Given 2 series

\sum_{n = 0}^{+ \infty} a_{n}, \sum_{n = 0}^{+ \infty} b_{n}

, then what’s

\sum_{n = 0}^{+ \infty} a_{n} b_{n}

? With analogy to polynomial, we have

(\sum_{n = 0}^{+ \infty} a_{n}) (\sum_{n = 0}^{+ \infty} b_{n}) \overset{?}{=} \sum_{n = 0}^{+ \infty} (\sum_{j = 0}^{+ \infty} a_{j} b_{n - j})

A figure to understand the formula above is

R.H.S. {\begin{matrix} \sum_{j = 0}^{0} a_{j} b_{0 - j} & a_{0} b_{0} & \to & a_{0} b_{1} & \to & a_{0} b_{2} & \dots & a_{0} \sum_{k = 0}^{+ \infty} b_{k} \\ ↙ & ↙ \\ \sum_{j = 0}^{1} a_{j} b_{1 - j} & a_{1} b_{0} & \to & a_{1} b_{1} & \to & a_{2} b_{2} & \dots & a_{1} \sum_{k = 0}^{+ \infty} b_{k} \\ ↙ & ↙ \\ \sum_{j = 0}^{2} a_{j} b_{2 - j} & a_{2} b_{0} & \to & a_{2} b_{1} & \to & a_{2} b_{2} & \dots & a_{2} \sum_{k = 0}^{+ \infty} b_{k} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix}} L.H.S.

We have

Theorem 5.2.18 Assuming $\sum_{n = 0}^{+ \infty} a_{n}, \sum_{n = 0}^{+ \infty} b_{n}$ are both absolutely convergent, then series $\sum_{n = 0}^{+ \infty} (\sum_{j = 0}^{+ \infty} a_{j} b_{n - j})$ is also absolutely convergent, and $(\sum_{n = 0}^{+ \infty} a_{n}) (\sum_{n = 0}^{+ \infty} b_{n}) = \sum_{n = 0}^{+ \infty} (\sum_{j = 0}^{+ \infty} a_{j} b_{n - j})$

Example 5.2.7 Assuming $x \in R, C$ or matrix, define $\exp x = \sum_{n = 0}^{+ \infty} \frac{x^{n}}{n!}$ , prove that when $x, y$ is commutable, $\exp (x + y) = \exp x \exp y$ .

Since $‖ \frac{x^{n}}{n!} ‖ \leq \frac{∥ x ∥^{n}}{n!} = c_{n}, \frac{c_{n + 1}}{c_{n}} = \frac{∥ x ∥}{n + 1} \to 0$ According to D’Alembert criterion, $\sum_{n = 0}^{+ \infty} \frac{x^{n}}{n!}$ is absolutely convergent. Therefore, $(\sum_{n = 0}^{+ \infty} \frac{x^{n}}{n!}) (\sum_{n = 0}^{+ \infty} \frac{y^{n}}{n!}) = \sum_{n = 0}^{+ \infty} \frac{1}{n!} \sum_{j = 0}^{n} \frac{n!}{j! (n - j)!} x^{j} y^{n - j} = \sum_{n = 0}^{+ \infty} \frac{(x + y)^{n}}{n!} = \exp (x + y)$

Assuming $\exp x$ is derivable with respect to $x$ , and $(\exp x)^{'}$ could be derived by taking the derivative term-by-term, then we have ${\begin{cases} (\exp x)^{'} = \sum_{n = 0}^{+ \infty} {(\frac{x^{n}}{n!})}^{'} = \exp x \\ \exp 0 = 1 \end{cases}$ Then $\exp x = e^{x}$