问题
填空题
数列{
|
答案
由题意可得an=
=1 n(n+1)
-1 n 1 n+1
∴Sn=
+1 1×2
+… +1 2×3 1 n(n+1)
=1-
+1 2
-1 2
+…+1 3
-1 n 1 n+1
=1-
=1 n+1 n n+1
∴
Sn=lim n→∞ lim n→∞
=n n+1 lim n→∞
=11 1+ 1 n
故答案为:1
数列{
|
由题意可得an=
=1 n(n+1)
-1 n 1 n+1
∴Sn=
+1 1×2
+… +1 2×3 1 n(n+1)
=1-
+1 2
-1 2
+…+1 3
-1 n 1 n+1
=1-
=1 n+1 n n+1
∴
Sn=lim n→∞ lim n→∞
=n n+1 lim n→∞
=11 1+ 1 n
故答案为:1