一道极限题··急求。。
\u4e00\u9053\u6c42\u6781\u9650\u7684\u9898\u5206\u6790\uff1a\u89c2\u5bdf\u77e5\u6b64\u9898\u6781\u9650\u7684\u7c7b\u578b\u4e3a1\u7684\u65e0\u7a77\u5927\u6b21\u65b9\uff0c
\u6240\u4ee5\u8981\u51d1\u6210lim\uff08x\u2192\u221e\uff09\uff08\uff081+1/x\uff09^x\u8fd9\u79cd\u5f62\u5f0f\uff0c
\u4e0b\u9762\u7684x\u7edf\u4e00\u6362\u6210n\u5c31\u53ef\u4ee5\u4e86\uff0c\u8fd8\u6709\u54ea\u4e00\u6b65\u4e0d\u660e\u767d\uff0c
\u53ef\u4ee5\u7ed9\u6211\u7559\u8a00\uff0c\u6211\u4e0a\u7ebf\u540e\u5e2e\u4f60\u89e3\u7b54\uff01
lim\uff08x\u2192\u221e\uff09[\uff082n+3\uff09/(2n+1)]^(n+1)
=lim\uff08x\u2192\u221e\uff09[\uff082n+1+2\uff09/(2n+1)]^(n+1)
=lim\uff08x\u2192\u221e\uff09(1+2/(2n+1))^(n+1)
=lim\uff08x\u2192\u221e\uff09[\uff081+2/(2n+1))^(2n+1)/2]^[2(n+1)/(2n+1)]
=lim\uff08x\u2192\u221e\uff09[\uff081+2/(2n+1))^(2n+1)/2]^lim\uff08x\u2192\u221e\uff09[2(n+1)/(2n+1)](\u6781\u9650\u8fd0\u7b97\u6cd5\u5219\uff1a\u5e42\u7684\u6781\u9650\u7b49\u4e8e\u6781\u9650\u7684\u5e42)
=e^lim\uff08x\u2192\u221e\uff09[1+1/(2n+1)]
=e^[lim\uff08x\u2192\u221e\uff091+lim\uff08x\u2192\u221e\uff091/(2n+1)]
=e^(1+0)
=e
\u6709\u7591\u95ee\uff0c\u53ef\u4ee5hi\u6211\u3002
\u70b9\u51fb\u770b\u5927\u56fe
{√a2-1}=1
{√a6-1}=3
因此每项之间差为(3-1)/4 =0.5
(√an-1) = (√a2-1) + 0.5 * (n-2) =0.5n
an = (0.5n)^2+1
a(n+2)-a(n+1) = (0.5n+1)^2-(0.5n+0.5)^2 = 0.5(n+1.5)
a(n+1)-a(n) = (0.5n+0.5)^2 -(0.5n)^2 = 0.5(n+0.5)
bn = 1/((n+3/2)(n+1/2)/4) = 16/[(2n+3)(2n+1)]
= 8[1/(2n+1) -1/(2n+3)]
Sn=b1+b2+b3+...+bn = 8[1/3-1/5+1/5-1/7+....+1/(2n+1) -1/(2n+3)]
=8[1/3-1/(2n+3)]
极限是8/3
bn=1/（an+2-an+1）（an+1-an） 是 bn=1/ [（an+2-an+1）（an+1-an）] 吗?
√(a6-1)=3 √(a2-1)=1
a'n=√(an-1)是等差数列，a'2=1,a'6=3
a'2=a'1+d=1
a'6=a'1+5d=3
a1'=1/2 d=1/2 √(a1-1)=1/2 a1=5/4
an'=(1/2)+(n-1)/2=n/2
a'n-1=(n-1)/2
an-an-1=[√(an-1)]^2-[√(an-1 -1)]^2=n^2/4-(n-1)^2/4=(2n-1)/4
Sn-a1-Sn-1=[(3/4)+(2n-1)/4](n-1)/2 an=5/4+(n-1)^2/4
an+2-an+1=(n+1)^2/4-n^2/4=(2n+1)/4
an+1-an=n^2/4-(n-1)^2/4=(2n-1)/4
bn=16/[(2n-1)(2n+1)]=8*[1/(2n-1)-1/(2n+1)]
b1=8*(1-1/3)
b1+b2+..+bn=8*[1-1/(2n+1)]
lim(n→+∞)b1+b2+b3+..+bn=8
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