∞ ∑ 1/n(n+1)(n+2) n=1 ∞ ∑ n的n+（1／n)次方除以n+（1／n)的n 次方 n=1

Question

这是两道题 ，第一道求和 。第二道判断收敛性 。前面的无穷和后面的n=1都应该在那个求和符号下面。。不好意思哈

Answer 1

解法一：
幂级数求和：
f(x)=∑ x^n/n(n+1)(n+2)
=∑(n+1-n)x^n/n(n+1)(n+2)
=∑x^n/n(n+2)-x^n/(n+1)(n+2)
=∑0.5x^n(1/n-1/(n+2))-(x^n/(n+1)-x^n/(n+2))
=0.5∑x^n/n-∑x^n/(n+1)+0.5∑x^n/(n+2)。。。1
其中：
A、∑x^n/n
=∫(0,x)(∑x^n/n)'dx
=∫(0,x)∑x^(n-1)dx
=∫(0,x)1/(1-x)dx
=-ln(1-x)
B、∑x^n/(n+1)
=(1/x)*∑x^(n+1)/(n+1)
=(1/x)*∫(0,x)(∑x^(n+1)/(n+1))'dx
=(1/x)*∫(0,x)∑x^ndx
=(1/x)*∫(0,x)(x/(1-x))dx
=(1/x)*(-x-ln(1-x))
=-1-(ln(1-x))/x
C、∑x^n/(n+2)
=(1/x^2)∑x^(n+2)/(n+2)
=(1/x^2)*∫(0,x)(∑x^(n+2)/(n+2))'dx
=(1/x^2)*∫(0,x)(∑x^(n+1)dx
=(1/x^2)*∫(0,x)x^2/(1-x)dx
=(1/x^2)*∫(0,x)-1-x+1/(1-x)dx
=(1/x^2)*(-x-0.5x^2-ln(1-x))
=-0.5-1/x-(ln(1-x))/x^2
将ABC代入“1”，
并令x=1,求出f(x)即可
代入得到：原式=0.5*0-(-1)+0.5*(-1.5)=0.25

解法二：
拆项相消法：
1/n(n+1)(n+2)=0.5(1/n(n+1)-1/(n+2)(n+1))
=>
Sn=∑0.5(1/n(n+1)-1/(n+2)(n+1))
=0.5(1/(1*2)-1/(n+1)(n+2))
n极限时，Sn=0.5*1/2=0.25

解法三：
待定系数法：
1/n(n+1)(n+2)=0.5/n-1/(n+1)+0.5/(n+2)
观察到三者系数之和为零（0.5-1+0.5）
累加可有消去性质
=>
Sn=(0.5*1-1/2+0.5*1/3)+(0.5*0.5-1/3+0.5*1/4)+(0.5*1/3-1/4+0.5*1/5)+...+(0.5/n-1/(n+1)+0.5/(n+2))
=0.25-1/(n+1)-0.5/(n+2)
n极限时
Sn=0.25

解法四：
公式法：
∑1/n(n+1)...(n+m)
=lim:(1/m)*(1/m!-1/(n+1)(n+2)...(n+m))
=1/(m*m!)
令m=2
=>
原式=1/4

由于分子分母都有幂次形式
故考虑用根值法判断敛散性：
lim:n√An
=lim:(An)^(1/n)
=lim:n^((n+1/n)/n)/(n+1/n)
=lim:n^(1+1/n^2)/(n+1/n)
=lim:n/(n+1/n)*lim:n^(1/n^2)/(n+1/n)
=lim:n^(1/n^2)/(n+1/n)
=0<1
=>
收敛

Answer 2

6646