5.14 1) 2 (√k+ 1−
√k) = 2·
(√k+ 1−
√k) (√k+ 1 +√ k)
√k+ 1 +√
k = 2·
(k+ 1)−k
√k+ 1 +√ k
= 2·
1
√k+ 1 +√ k
<2· 1
√k+√
k = 2· 1 2√
k = 1
√k
2)
n
X
k=1
1
√k
>
n
X
k=1
2 (√
k+ 1−
√k) = 2
n
X
k=1
√k+ 1−
√k
= 2 (√ 2−
√1
| {z }
k=1
+√ 3−
√2
| {z }
k=2
+√ 4−
√3
| {z }
k=3
+. . .+√
n+ 1−
√n
| {z }
k=n
)
= 2 (−
√1 +√
n+ 1) = 2 (√
n+ 1−1)
3) Puisque la suite des sommes partielles est non bornée, elle diverge.
Analyse : séries Corrigé 5.14