Question

高等数学，如图，这个过程为什么不对？

正确答案2-2ln2

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Answer 1

1/(1-t) = 1+t+t^2+.....
∫(0->x) dt/(1-t) = ∫(0->x) [1+t+t^2+.....] dt
-ln(1-x) = x+x^2/2 +....
x= 1/2
-ln(1/2) = 1/2+1/(2.2^2) +1/(3.2^3)+....
ln2 =1/2+1/(2.2^2) +1/(3.2^3)+....
ln2 - 1/2 =1/(2.2^2) +1/(3.2^3)+....
ln2 - 1/2 =lim(n->无穷) ∑(i:1->n) [1/(i+1)] [1/2^(i+1) ]
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lim(n->无穷) ∑(i:1->n) i/[(i+1).2^i ]
=lim(n->无穷) ∑(i:1->n) [ 1- 1/(i+1)] (1/2^i)

=lim(n->无穷) ∑(i:1->n) (1/2^i) -lim(n->无穷) ∑(i:1->n) [1/(i+1)] (1/2^i)
= 1 - lim(n->无穷) ∑(i:1->n) [1/(i+1)] (1/2^i)
= 1 - 2lim(n->无穷) ∑(i:1->n) [1/(i+1)] [1/2^(i+1) ]
=1 - 2[ln2 - 1/2 ]
=2-2ln2

Answer 2

错在“∑x^n=1/(1-x)”。∵n=1，∴ ∑x^n=x/(1-x)。再令x=1/2即可。