幂级数求和：0到正无穷x^n/(n+1)怎么作,

令a_n = x^n/(n+1).严格来讲,这个题解法如下
(1)确定级数收敛域
用比值判别法
|a_{n+1}/a_n| = |x|(n+1)/n+2 -> |x| (n ->∞).因此当|x| 1时,级数发散.当x = -1时,级数为交错调和级数,收敛.当x = 1时,级数为调和级数,发散.故收敛域为-1 ≤ x (2)根据定义,任意实数的0次幂等于1,0也不例外,即0^0 = 1.对于n > 0,有0^n = 0.显然S(0)＝1.
令Sn(x)为前n项和,当x在[-1,1)内且x≠0时,有
Sn(x) = 1/x Σ{k=0,n} x^(k+1)/(k+1)
记fn(x) = Σ{k=1,n} x^(k+1)/(k+1),则Sn(x) = (fn(x)+x)/x.
dfn(x)/dx = Σ{k=1,n} d[x^(k+1)/(k+1)]/dx
= Σ{k=1,n} x^(k-1)
= (1-x^n)/(1-x).
可见fn'(x) = (1-x^n)/[x(1-x)],根据牛顿-莱布尼兹公式,有
fn(x) - fn(0) = ∫{0,x} (1-t^n)/(1-t) dt
= ∫{0,x}1/(1-t) dt - ∫{0,x}t^n/(1-t) dt
= ln(1-x) - ∫{0,x} t^n/(1-t) dt
从而
Sn(x) = (ln(1-x) - ∫{0,x} t^n/(1-t) dt + fn(0) + x)/x= ln(1-x)/x + 1 + fn(0)/x - (1/x) ∫{0,x}t^n/(1-t) dt.易知对任意n>0,fn(0) = 0,故Sn(x) = ln(1-x)/x + 1 - (1/x) ∫{0,x} t^n/(1-t) dt.从而
S(x) = lim{n->∞} Sn(x)
=lim{n->∞} [ ln(1-x)/x + 1 - (1/x) ∫{0,x} t^n/(1-t) dt]
= ln(1-x)/x + 1 - lim{n->∞}∫{0,x} t^n/(1-t) dt.
注意到当-1∞).同时当x = -1时,也可证明∫{0,x} t^n/(1-t) dt -> 0 (n->∞)(较复杂).
故x在[-1,1)内且x≠0时,S(x) = ln(1-x)/x + 1,而S(0)＝1.