呼叫高数大师!求指导!
1个回答
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考察一般项:
1/[2(1+2+...+k)]=1/[2k(k+1/2]=1/[k(k+1)]=1/k -1/(k+1)
n
∑ 1/[2(1+2+...+k)]ⁿ
k=1
=[1/1 -1/2 +1/2 -1/3+...+ 1/n -1/(n+1)]ⁿ
=[1 -1/(n+1)]ⁿ
={[1+1/(-n-1)]⁻ⁿ⁻¹}⁻¹ /[1+ 1/(-n-1)]
={[1+1/(-n-1)]⁻ⁿ⁻¹}⁻¹ /[1- 1/(n+1)]
I=limxn
x→∞
=lim {[1+1/(-n-1)]⁻ⁿ⁻¹}⁻¹ /[1- 1/(n+1)]
x→∞
=e⁻¹/1
=1/e
1/[2(1+2+...+k)]=1/[2k(k+1/2]=1/[k(k+1)]=1/k -1/(k+1)
n
∑ 1/[2(1+2+...+k)]ⁿ
k=1
=[1/1 -1/2 +1/2 -1/3+...+ 1/n -1/(n+1)]ⁿ
=[1 -1/(n+1)]ⁿ
={[1+1/(-n-1)]⁻ⁿ⁻¹}⁻¹ /[1+ 1/(-n-1)]
={[1+1/(-n-1)]⁻ⁿ⁻¹}⁻¹ /[1- 1/(n+1)]
I=limxn
x→∞
=lim {[1+1/(-n-1)]⁻ⁿ⁻¹}⁻¹ /[1- 1/(n+1)]
x→∞
=e⁻¹/1
=1/e
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