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3. Si un ne tend pas vers 0 quand n tend vers +∞, la série un est divergente. Supposons que lim u n = 0. Alors un ∼ − ln(1 − un ), ce qui s’écrit −un ∼ ln(Dn−1 /Dn ). Or − n ln(1 − uk ) = − ln D0 + ln Dn → +∞ k=1 si n → +∞. Par suite, la série Pour n un diverge. 1, on a : vn = Dn − Dn−1 Dnα Dn Dn−1 dt · tα Comme α > 1, pour tout N ∈ N, on a donc +∞ N vn n=1 ce qui montre que la série d0 dt < +∞, tα vn est convergente. 4. 1. Si n est assez grand, Rn−1 < 1. Si a grand. D’où la convergence de vn .

Soit n ap bq xpq . Sn (x) = p,q=1 On va prouver que : lim Sn (x) = F (x) = G(x). n Par symétrie, il suffit de prouver la première égalité. Soit ε > 0. Il existe n0 ∈ N tel que : n n0 ⇒ ∞ ∞ |bp |r p < ε , p=n+1 |ap |r p < ε. p=n+1 On a : ∞ s=1 ∞ bs f (xs ) − Sn (x) n |bs | |f (xs )| + s=n+1 ∞ A bs f (xs ) − s=1 |bs |r s + s=n+1 n Aε + s=1 |bs | n q=1 ∞ |bs |. s=1 ∞ |aq |r sq q=n+1 rs q r q |aq | q=n+1 n r aq xsq Solutions des exercices Comme r s 49 r < r < 1, on a donc : ∞ ∞ bs f (xs ) − Sn (x) s=1 car ∞ |bs |ε Aε + s=1 rs r r q |aq | q=n+1 ∞ rs r q rs r rs · r |aq |r q ε |bs |r s Aε + Bε.

La photocopie non autorisée est un délit. 1. Une série de fonctions gente sur X s’il existe une série conditions suivantes : (i) La série µn est convergente. (ii) Pour tout n ∈ N, on a sup{|fn (x)| ; x ∈ X} µn . 2. Si une série de fonctions sur une partie X de K est normalement convergente sur X , elle est uniformément convergente sur X . Démonstration. C’est immédiat d’après le critère de Cauchy uniforme. 3. Soient g = (gn )n et h = (hn )n des suites de fonctions sur X vérifiant les conditions suivantes : (i) Pour tout x ∈ X , la suite gn (x) n est à termes réels et décroissante.

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