matlab级数求和：∑(1+1/2+...+1/n)/(n+1)(n+2)

这个级数可以通过分解分式、拆项、变形等方法来求和。

首先，将每一项分式进行拆分，得到：

(1+1/2+...+1/n)/(n+1)(n+2) = [(1/(n+1)) + (1/(n+1) + 1/(n+2)) + ... + (1/(n+1) + 1/(n+2) + ... + 1/n)]/(n+1)(n+2)

接着，将分式中的每一项进行拆分，得到：

1/(n+1) + 1/(n+1) + 1/(n+2) + ... + 1/(n+1) + 1/(n+2) + ... + 1/n

= [(n-1)/(n+1) + (n-2)/(n+2) + ... + 1/2 + 1/2] + [1/(n+1) + 1/(n+2) + ... + 1/n]

= [(n-1)/(n+1) + (n-2)/(n+2) + ... + 1/2] + [1/(n+1) + 1/(n+2) + ... + 1/n + 1/2]

= [n/(n+1) - 1/(n+1) + (n-1)/(n+2) - 1/(n+2) + ... + 2/3 - 1/3 + 1/2]/2 + [1/(n+1) + 1/(n+2) + ... + 1/n + 1/2]

= [(n-1)/2n - 1/(n+1) + (n-3)/2(n+1) - 1/(n+2) + ... + 1/6 - 1/3 + 1/2]/2 + [1/(n+1) + 1/(n+2) + ... + 1/n + 1/2]

= [(n-1)/2n - 1/(n+1) + (n-3)/2(n+1) - 1/(n+2) + ... + 1/6 - 1/3 + 1/2]/2 + [H_n - H_{n+1} + 1/2]

其中，H_n 表示调和级数，即 H_n = 1 + 1/2 + ... + 1/n。

将上式代入原级数中，得到：

∑(1+1/2+...+1/n)/(n+1)(n+2)

= [(n-1)/2n - 1/(n+1) + (n-3)/2(n+1) - 1/(n+2) + ... + 1/6 - 1/3 + 1/2]/2 + [H_n - H_{n+1} + 1/2]/[(n+1)(n+2)]

= [n/(2n+2) - H_{n+1}/(n+1) + 1/2]/2 + [H_n - H_{n+1} + 1/2]/[(n+1)(n+2)]

= [(n+1)H_n - nH_{n+1}]/[(n+1)(n+2)]

因此，该级数的和为 [(n+1)H_n - nH_{n+1}]/[(n+1)(n+2)]。