问题
填空题
设f(n)=
|
答案
由题意可得,f(n+1)-f(n)=(
+1 n+2
+1 n+3
+…+1 n+4
)-(1 2n+2
+1 n+1
+1 n+2
+…+1 n+3
)=1 2n
+1 2n+1
-1 2n+2
,1 n+1
则
n2[f(n+1)-f(n)]=lim n→+∞
n2(lim n→+∞
+1 2n+1
-1 2n+2
)=1 n+1
(n2•lim n→+∞
)=1 (2n+1)(2n+2)
(lim n→+∞
)=n2 4n2+6n+2
(lim n→+∞
)=1 4+
+6 n 2 n2
=1 4+0+0
,1 4
故答案为
.1 4