Products of absolutely convergent

Let ∞ ∞ X X ak, bl k=0 l=0 be two series. Define for n ∈ N n X X cn := akbl = akbn−k = a0bn + a1bn−1 + ... + an−1b1 + anb0. k+l=n k=0 P∞ P∞ The series n=0 cn is called the of the series k=0 ak and P∞ l=0 bl. Motivation: “multiplying termwise and summing up:”, as in finite sums:

a0 a1 a2 a3 ... b0 a0b0 a1b0 a2b0 a3b0 ... b1 a0b1 a1b1 a2b1 ... .. b2 a0b2 a1b2 . . b3 a0b3 . . . . .

The term cn contains the products on the n-th rising diagonal in this table. Proposition 1: Assume that

∞ ∞ X X ak = A, bl = B k=0 l=0 are absolutely convergent. Then the Cauchy product is absolutely convergent as well, and ∞ X cn = AB. n=0 P Proof: I. of cn: PN Let σN := n=0 |cn|. So the (σN ) is increasing. Moreover,

N X X X X σN ≤ |ak||bl| = |ak||bl| ≤ |ak||bl| n=0 k+l=n k+l≤N k≤N, l≤N N N ∞ ∞ X X X X = |ak| |bl| ≤ |ak| |bl|. k=0 l=0 k=0 l=0 P P So (σN ) bounded above ⇒ |cn| convergent ⇒ cn absolutely convergent.

1 II. Convergence to AB: Define the of index pairs

IN := {(k, l) | k + l ≤ 2N ∧ (k > N ∨ l > N)}.

Then

2N N N X X X X cn − ak bl = akbl

n=0 k=0 l=0 (k,l)∈IN 2N 2N N N X X X X X N→∞ ≤ |ak||bl| = |ak| |bl| − |ak| |bl| −→ 0.

(k,l)∈IN k=0 l=0 k=0 l=0

So ∞ 2N N N ! X X X X cn = lim cn = lim ak bl = AB. N→∞ N→∞ n=0 n=0 k=0 l=0

The exponential

Lemma: For all x ∈ R, the series

∞ X xk k! k=0 is absolutely convergent. xk Proof: Without loss of generality x 6= 0. Let ak := k! . Then |a | |x| k+1 = k−→→∞ 0. |ak| k P P : |ak| convergent ⇒ ak absolutely convergent. Definition: Let the function exp : R −→ R be given by

∞ X xk exp(x) := . k! k=0 Proposition 2: I. exp(1) = e

II. exp(x + y) = exp(x) exp(y) for all x, y ∈ R. Proof: I. Let n n  1  X xk e := 1 + , e0 := . n n n k! k=0

2 0 By definition: en → e, en → exp(1) als n → ∞. a) :

n n n  1  X n 1 X 1 n · (n − 1) · ... · (n − k + 1) e = 1 + = = 1 + n n k nk k! n · ... · n k=0 k=1 n X 1 ≤ = e0 . k! n k=0 Limit n → ∞: e ≤ exp(1). b) Choose m ∈ N+ en n ≥ m arbitrary. Then n X 1 n · (n − 1) · ... · (n − k + 1) e = 1 + n k! nk k=1 m X 1 n · (n − 1) · ... · (n − k + 1) ≥ 1 + k! nk k=1 m X 1 n − 1 n − 2 n − k + 1 = 1 + · 1 · · · ... · . k! n n n k=1 m fixed, n → ∞: m X 1 e ≥ 1 + = e0 . k! m k=1 Now m → ∞: e ≥ exp(1). a) and b) together imply I. II. The series ∞ ∞ X xk X yl exp(x) = , exp(y) = k! l! k=0 l=0 are absolutely convergent, according to the lemma. Proposition 1:

∞ X exp(x) exp(y) = cn, n=0 where n n X xk yn−k 1 X n 1 c = = xkyn−k = (x + y)n. n k! (n − k)! n! k n! k=0 k=0 So ∞ X (x + y)n exp(x) exp(y) = = exp(x + y). n! n=0

Reminder: If a > 0 and p ∈ Q met p = m/n, m ∈ Z, n ∈ N+, then by definition √ ap := n am.

3 Proposition 3: For all rational numbers p and all x ∈ R we have exp(px) = exp(x)p.

Proof: I. ∀x ∈ R, n ∈ N : exp(nx) = exp(x)n. Proof by induction: Statement is true for n = 0 (Check!). Assume exp(nx) = exp(x)n for an n ∈ N. Then exp((n + 1)x) = exp(nx + x) = exp(nx) exp(x) = exp(x)n exp(x) = exp(x)n+1.

II. exp(−x) = exp(x)−1 because of

exp(x) exp(−x) = exp(0) = 1.

1/n III. ∀x ∈ R, n ∈ N+ : exp(x/n) = exp(x) because of  x   x n exp(x) = exp n =I exp . n n

IV. Let now in general p = m/n with m ∈ Z, n ∈ N+. Without loss of generality p 6= 0. Distinguish two cases: a) m > 0: m   x m  m exp x =I exp III= exp(x)1/n = exp(x)m/n. n n b) m < 0

−1 m   −m  −m  a)  −1 exp x = exp − x =II exp x = exp(x)−m/n = exp(x)m/n. n n n

Corollary: For all x ∈ Q we have exp(x) = ex.

(Check!)

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