作业帮微积分：求下例数列的极限,谢谢.
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微积分：求下例数列的极限,谢谢.
微积分：求下例数列的极限,谢谢.
微积分：求下例数列的极限,谢谢.
1+1/2+1/4+...+1/2^n = [1-1/2^(n+1)]/(1-1/2) = 2[1-1/2^(n+1)],
1+1/3+1/9+...+1/3^n = [1-1/3^(n+1)]/(1-1/3) = (3/2)[1-1/3^(n+1)],
lim_{n->无穷}[1+1/2+1/4+...+1/2^n]/[1+1/3+1/9+...+1/3^n] = 2[1-0]/{(3/2)[1-0]} = 4/3.
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1+2+...+n = n(n+1)/2,
[1+2+...+n]/(n+2) - n/2 = n(n+1)/[2(n+2)] - n/2 = (n/2)[(n+1)/(n+2)-1] = -n/[2(n+2)] = (-1/2)/[1+2/n],
lim_{n->无穷}[(1+2+...+n)/(n+2)-n/2] = (-1/2)/[1+0] = -1/2.
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1-1/k = (k-1)/k.
(1-1/2)(1-1/3)(1-1/4)...[1-1/(n-1)][1-1/n] = (2-1)/2*(3-1)/3*(4-1)/4...(n-1-1)/(n-1)*(n-1)/n
= (2-1)/n = 1/n,
lim_{n->无穷}(1-1/2)(1-1/3)...(1-1/n) = 0.
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1/[(2n-1)(2n+1)] = (1/2)[1/(2n-1) - 1/(2n+1)],
1/(1*3) + 1/(3*5) + ...+ 1/[(2n-3)(2n-1)] + 1/[(2n-1)(2n+1)]
= (1/2)[1/1 - 1/3 + 1/3 - 1/5 + ...+ 1/(2n-3) - 1/(2n-1) + 1/(2n-1) - 1/(2n+1)]
= (1/2)[1/1 - 1/(2n+1)]
= (1/2)[2n/(2n+1)]
= n/(2n+1)
= 1/(2+1/n)
lim_{n->无穷}[1/(1*3) + 1/(3*5)+...+1/[(2n-1)(2n+1)]] = 1/(2+0) = 1/2,