Ngô Quốc Anh

August 28, 2007

Sử dụng Tích phân xác định tính tổng vô hạn

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 16:15

Đề bài: Chứng minh hệ thức

\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\;\frac{1}{m^{2}n+2mn+mn^{2}}\;\;=\;\;\boxed{\frac{7}{4}}

Lời giải:

More generally, let k be a natural number. Then

\sum_{m_{1},m_{2},\cdots, m_{n}\geq 1}\frac{1}{m_{1}\cdots m_{n}(m_{1}+m_{2}+\cdots+m_{n}+k)}=

            =(-1)^{n}\int_{0}^{1}x^{k-1}\ln^{n}(1-x)dx.

We have that

S=\sum\frac{1}{m_{1}\cdots m_{n}}\int_{0}^{1}x^{m_{1}+\cdots+m_{n}+k-1}dx

=\int_{0}^{1}x^{k-1}\sum_{m_{1}}\frac{x^{m_{1}}}{m_{1}}\cdots\sum_{m_{n}}\frac{x^{m_{n}}}{m_{n}}dx      

=\int_{0}^{1}x^{k-1}\left(\ln\frac{1}{1-x}\right)^{n}dx                   

=\int_{0}^{1}(-1)^{n}(1-x)^{k-1}\ln^{n}(x)dx.           

When n=2 and k=2 we get that the sum equals

\int_{0}^{1}(1-x)\ln^{2}(x)dx=\frac{7}{4}

.

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