Tutorato di Analisi 3 - Dipartimento di Matematica

+∞∑
arctan ( ) ∣
k ∣∣∣∣
2
∣
+∞∑
k ∣∣∣∣
2
n
=
2 n
∣ k 2 ≤
2
∣ k 2 = 1 ∑
+∞
n 4 k=1
k=1
k=1
+∞∑
anche in l 1 perché ‖x n − x‖ 1 =
∣
k=1
(
+∞∑ ∣ arctan k
)∣ ∣ n∑ k +∞
n
+
2 n
k 2 ≤
2
k 2 +
∑
k=n+1
∑
n
≤ 1 n 2
k=1
1 + π 2
+∞∑
k=n+1
k=1
1
k 2 = 1 n + π 2
1 n→+∞
n→+∞
k 2 → 0; inoltre, x n → x
arctan ( k
n 2 )
k 2 ∣ ∣∣∣∣
=
k=n+1
+∞∑
k=n+1
π
2
k 2 = 1 n 2
n∑
k=1
∑ n
k=1
1
k 2 n→∞
→ 0.
∣
∣arctan ( k
n 2 )∣ ∣
k 2 +
1
k + π 2
+∞∑
k=n+1
1
k 2 ≤