By H. A Lauwerier

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2 ∞ ak converge, vale la condizione necessaria lim ak = 0. 8. Poich´e la serie k=1 Pertanto lim k→∞ 1 non pu` o valere 0 e dunque la serie ak k→∞ ∞ k=1 1 non pu`o convergere. ak 9. Studio della convergenza di serie a termini positivi: a) Converge. b) Osserviamo che il termine generale ak tende a +∞ per k → ∞. 6 la serie diverge positivamente. 15. 14: ak+1 3k+1 k! = lim ; k→∞ ak k→∞ (k + 1)! 3k lim scrivendo (k + 1)! = (k + 1)k! e sempliﬁcando, si ottiene lim k→∞ ak+1 3 = 0. = lim k→∞ k + 1 ak Ne segue che la serie converge.

Sia A l’insieme di convergenza della serie ak xk . k=0 Se A = {0}, si ha il caso a). Se A = R, si ha il caso b). 26, la serie converge puntualmente e assolutamente per ogni x ∈ R. Per quanto riguarda la convergenza uniforme su un intervallo [a, b], poniamo L = max(|a|, |b|). 20 con Mk = |ak |Lk . Si supponga ora che A contenga punti diversi da 0 ma che non sia l’intera retta. 26, A non pu` o contenere alcun punto x con |x| > |x|, quindi l’insieme A `e limitato. Poniamo R = sup A; si ha R > 0, perch´e A non si riduce al solo {0}.

K=k0 a Si usa dire che la serie `e integrabile termine a termine. 19 (Derivazione per serie) Sia {fk }k≥k0 una successione di funzioni di classe C 1 su un intervallo I = [a, b]. Supponiamo che esistano due funzioni s e t deﬁnite su I tali che ∞ i) fk (x) = s(x) , ∀x ∈ I; fk (x) = t(x) , ∀x ∈ I e la convergenza sia uniforme su I. k=k0 ∞ ii) k=k0 ∞ Allora s ∈ C 1 (I) e si ha s = t. Inoltre fk converge uniformemente a s k=k0 ∞ fk converge uniformemente a s ). su I (e k=k0 In altre parole, il teorema aﬀerma che ∞ ∞ fk (x) = k=k0 fk (x) , ∀x ∈ I .