令a_n = x^n/(n+1).嚴格來講,這個題解法如下
(1)確定級數(shù)收斂域
用比值判別法
|a_{n+1}/a_n| = |x|(n+1)/n+2 -> |x| (n ->∞).因此當|x| 1時,級數(shù)發(fā)散.當x = -1時,級數(shù)為交錯調(diào)和級數(shù),收斂.當x = 1時,級數(shù)為調(diào)和級數(shù),發(fā)散.故收斂域為-1 ≤ x < 1.
(2)根據(jù)定義,任意實數(shù)的0次冪等于1,0也不例外,即0^0 = 1.對于n > 0,有0^n = 0.顯然S(0)＝1.
令Sn(x)為前n項和,當x在[-1,1)內(nèi)且x≠0時,有
Sn(x) = 1/x Σ{k=0,n} x^(k+1)/(k+1)
記fn(x) = Σ{k=1,n} x^(k+1)/(k+1),則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).
可見fn'(x) = (1-x^n)/[x(1-x)],根據(jù)牛頓-萊布尼茲公式,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
從而
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.易知對任意n>0,fn(0) = 0,故Sn(x) = ln(1-x)/x + 1 - (1/x) ∫{0,x} t^n/(1-t) dt.從而
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.
注意到當-1∞).同時當x = -1時,也可證明∫{0,x} t^n/(1-t) dt -> 0 (n->∞)(較復雜).
故x在[-1,1)內(nèi)且x≠0時,S(x) = ln(1-x)/x + 1,而S(0)＝1.