求极限,数列An是等差数列,且A1不等于0,Sn是前N项和求（n*（an））/Sn的极限和（Sn+S（n1））/（Sn+S（n-1））
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求极限,数列An是等差数列,且A1不等于0,Sn是前N项和求（n*（an））/Sn的极限和（Sn+S（n1））/（Sn+S（n-1））
求极限,数列An是等差数列,且A1不等于0,Sn是前N项和
求（n*（an））/Sn的极限
和（Sn+S（n1））/（Sn+S（n-1））
求极限,数列An是等差数列,且A1不等于0,Sn是前N项和求（n*（an））/Sn的极限和（Sn+S（n1））/（Sn+S（n-1））
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an = a1 + (n - 1)d (n = 1,2,3 ...)
Sn = (2*a1 + (n - 1)d) * n / 2 (x)
Therefore, n * an / Sn = 2 * (n^2 * d - n*d + n * a1) / (2 * a1 + n^2 * d - n * d), and its limit when n approaches infini...
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an = a1 + (n - 1)d (n = 1,2,3 ...)
Sn = (2*a1 + (n - 1)d) * n / 2 (x)
Therefore, n * an / Sn = 2 * (n^2 * d - n*d + n * a1) / (2 * a1 + n^2 * d - n * d), and its limit when n approaches infinity is 2.
Q2: Applying equation (x), we have （Sn+S（n+1））/（Sn+S（n-1））= (4 * a1 + 2n^2 * d ) / (4 * a1 + 2n^2 * d - 4nd + 2d). And its limit is 1 when n approaches infinity
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