幂级数求和:0到正无穷x^n/(n+1)怎么作,其中S(0)=1不懂怎么出来的,当n=0,x=0时和的第一项变为0^0/(0+1)=0吧,无意义,还是对求和的展开式误解,文科生请多多帮忙,虚心接受
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![幂级数求和:0到正无穷x^n/(n+1)怎么作,其中S(0)=1不懂怎么出来的,当n=0,x=0时和的第一项变为0^0/(0+1)=0吧,无意义,还是对求和的展开式误解,文科生请多多帮忙,虚心接受](/uploads/image/z/840112-16-2.jpg?t=%E5%B9%82%E7%BA%A7%E6%95%B0%E6%B1%82%E5%92%8C%EF%BC%9A0%E5%88%B0%E6%AD%A3%E6%97%A0%E7%A9%B7x%5En%2F%28n%2B1%29%E6%80%8E%E4%B9%88%E4%BD%9C%2C%E5%85%B6%E4%B8%ADS%EF%BC%880%EF%BC%89%EF%BC%9D1%E4%B8%8D%E6%87%82%E6%80%8E%E4%B9%88%E5%87%BA%E6%9D%A5%E7%9A%84%2C%E5%BD%93n%EF%BC%9D0%2Cx%3D0%E6%97%B6%E5%92%8C%E7%9A%84%E7%AC%AC%E4%B8%80%E9%A1%B9%E5%8F%98%E4%B8%BA0%5E0%2F%280%2B1%29%3D0%E5%90%A7%2C%E6%97%A0%E6%84%8F%E4%B9%89%2C%E8%BF%98%E6%98%AF%E5%AF%B9%E6%B1%82%E5%92%8C%E7%9A%84%E5%B1%95%E5%BC%80%E5%BC%8F%E8%AF%AF%E8%A7%A3%2C%E6%96%87%E7%A7%91%E7%94%9F%E8%AF%B7%E5%A4%9A%E5%A4%9A%E5%B8%AE%E5%BF%99%2C%E8%99%9A%E5%BF%83%E6%8E%A5%E5%8F%97)
幂级数求和:0到正无穷x^n/(n+1)怎么作,其中S(0)=1不懂怎么出来的,当n=0,x=0时和的第一项变为0^0/(0+1)=0吧,无意义,还是对求和的展开式误解,文科生请多多帮忙,虚心接受
幂级数求和:0到正无穷x^n/(n+1)怎么作,
其中S(0)=1不懂怎么出来的,当n=0,x=0时和的第一项变为0^0/(0+1)=0吧,无意义,还是对求和的展开式误解,文科生请多多帮忙,虚心接受
幂级数求和:0到正无穷x^n/(n+1)怎么作,其中S(0)=1不懂怎么出来的,当n=0,x=0时和的第一项变为0^0/(0+1)=0吧,无意义,还是对求和的展开式误解,文科生请多多帮忙,虚心接受
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 < 1.
(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.