Series

5.1 Series

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Series is a summation according to the order of the sequence.

Definition 5.1.1 Assuming $V$ is a linear space, ${a_{n}}_{n \geq 1}, a_{n} \in V, \forall n \geq 1$ , then $\sum_{n = 1}^{+ \infty} a_{n}$ is called a series.

Definition 5.1.2 Assuming the partial sum sequence (also sum of first $N$ terms) is $S_{N} = \sum_{n = 1}^{N} a_{n} \in V$ , $∥ \cdot ∥$ is a norm on $V$ , then series $\sum_{n = 1}^{+ \infty} a_{n}$ is convergent if $\exists S \in V$ such that $lim_{N \to + \infty} S_{N} \to S$ i.e. $\forall ϵ > 0$ , $\exists N_{ϵ}$ such that $\forall N \geq N_{ϵ}$ , $∥ S - S_{N} ∥ < ϵ$ . At this time $S$ is called the sum of series $\sum_{n = 1}^{+ \infty} a_{n}$ .

Definition 5.1.3 Series $\sum_{n = 1}^{+ \infty} a_{n}$ is divergent if it’s not convergent.

Note

When $V = R$ or $C$ , $\sum_{n = 1}^{+ \infty} a_{n}$ is a number series.
Assuming $V = C [0, 1]$ , it could be proven that $∥ f ∥_{\infty} = max_{0 \leq x \leq 1} | f (x) |$ is a norm on $V$ . Assuming $f_{n} \in C [0, 1]$ , if $\sum_{n = 1}^{+ \infty} f_{n}$ is convergent under $∥ \cdot ∥_{\infty}$ , then it’s uniformly convergent on $[0, 1]$ .
Assuming $V = R [0, 1]$ , define the norm
$∘ ∫-1---------- 2 ∥f∥2 = 0 |f(x)| dx$
and the inner product
$∫ 1 ⟨f, g⟩ = f (x)g(x)dx 0$
If $\sum_{n = 1}^{+ \infty} f_{n}$ is convergent under $∥ \cdot ∥_{2}$ , then it’s convergent in the mean square on $[0, 1]$ .

The sum of series.

Example 5.1.1 Geometric series $1 + x + x^{2} + \dots$ . When $∥ x ∥ < 1$ , it’s convergent, and

1 + x + x2 + ⋅⋅⋅ = (1 − x )−1

If $x \in R$ or $C$ , $∥ x ∥$ is the absolute value $| x |$ ; if $x$ is a square matrix, $∥ x ∥$ is the norm of matrix.

Proof. $\begin{aligned} S_{N} = 1 + x + \dots + x^{N - 1} \\ (1 - x) S_{N} = 1 + x + \dots + x^{N - 1} - (x + x^{2} + \dots + x^{N - 1} + x^{N}) = 1 - x^{N} \\ ∥ (1 - x) S_{N} - 1 ∥ = ∥ - x^{N} ∥ \leq ∥ x ∥^{N} \to 0, N \to + \infty \\ ∥ S_{N} - (1 - x)^{- 1} ∥ \leq ∥ 1 - x^{- 1} ∥ ∥ (1 - x) S_{N} - 1 ∥ \to 0 \end{aligned}$

Therefore, $S_{N} = (1 - x)^{- 1}$ .

Example 5.1.2 $a_{n} = \frac{1}{n (n + 1)}$ , $n \geq 1$ . We have

1 1 1 1 a1 + ⋅⋅⋅ + aN = ∖lef t(1 − 2-∖right) + ⋅⋅⋅ + ∖left(N-− N--+-1 ∖right) = 1 − N-+--1 → 1