Ngô Quốc Anh

September 30, 2007

Tính tổng của 1 chuỗi

Filed under: Các Bài Tập Nhỏ, Giải Tích 4 — Ngô Quốc Anh @ 6:49

Tính

\displaystyle S = \sum_{n=0}^{+ \infty}\frac{(3n)!}{(3n+3)!} = \sum_{n=0}^{+ \infty}\frac{1}{(3n+1)(3n+2)(3n+3)}.

Lời giải. Làm như sau

\displaystyle\begin{gathered}S = \frac{1}{{1\cdot2\cdot3}} + \frac{1}{{4\cdot5\cdot6}} + \frac{1}{{7\cdot8\cdot9}} +\cdots\hfill \\ \quad= \sum\limits_{n = 0}^\infty{\frac{1}{{(3n + 1)(3n + 2)(3n + 3)}}}\hfill \\ \quad= \frac{1}{2}\sum\limits_{n = 0}^\infty{\left( {\frac{1}{{3n + 1}} - \frac{2}{{3n + 2}} + \frac{1}{{3n + 3}}} \right)}\hfill \\ \quad= \frac{1}{2}\sum\limits_{n = 0}^\infty{\left( {\left[ {\frac{1}{{3n + 1}} - \frac{1}{{3n + 2}}} \right] + \left[ { - \frac{1}{{3n + 2}} + \frac{1}{{3n + 3}}} \right]} \right)}\hfill \\ \quad= \frac{1}{2}\left( {\left[ {1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{5} + \frac{1}{7} - \frac{1}{8} +\cdots } \right] + \left[ { - \frac{1}{2} + \frac{1}{3} - \frac{1}{5} + \frac{1}{6} - \frac{1}{8} + \frac{1}{9} -\cdots } \right]} \right) \hfill \\ \quad= \frac{1}{2}\left( {\left[ {\int_0^1 {\frac{{1 - x}}{{1 - {x^3}}}} dx} \right] + \left[ {\int_0^1 {\frac{{ - x + {x^2}}}{{1 - {x^3}}}} dx} \right]} \right) \hfill \\ \quad= \frac{1}{2}\left( {\int_0^1 {\frac{{1 - x}}{{1 + x + {x^2}}}} dx} \right) \hfill \\ \quad= \frac{1}{2}\left( {\frac{\pi }{{2\sqrt 3 }} - \frac{1}{2}\ln 3} \right). \hfill \\ \end{gathered}

Bài tập tương tự: Tính

\displaystyle S=\left(\frac{0!}{3!}\right)^2+\left(\frac{3!}{6!}\right)^2+\left(\frac{6!}{9!}\right)^2+\cdots.

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