ผลรวม

[Sabre]

จงหาค่าของ $\sum_{n = 1}^{\infty} \frac{n}{(2n-1)(2n+1)(2n+3)}$

[Amankris]

$\displaystyle\sum_{n=1}^{k}\dfrac{n}{(2n-1)(2n+1)(2n+3)}=\dfrac{k(k+1)}{2(2k+1)(2k+3)}$

[Sabre]

ขอบคุณครับ แต่ทำไมถึงเป็นแบบนั้นได้ครับ ช่วยบอกที

[yellow]

$\displaystyle\dfrac{1}{1\times 3\times 5}+\dfrac{2}{3\times 5\times 7}+\dfrac{3}{5\times 7\times 9}+...+\dfrac{k}{(2k-1)(2k+1)(2k+3)}$

$\displaystyle \frac{1}{4} (\frac{1}{1\times 3} - \frac{1}{3\times 5}) + \frac{2}{4} (\frac{1}{3\times 5} - \frac{1}{5\times 7}) + \frac{3}{4} (\frac{1}{5\times 7} - \frac{1}{7\times 9}) + ... + \frac{k}{4} (\frac{1}{(2k-1)(2k+1)} - \frac{1}{(2k+1)(2k+3)}) $

$\displaystyle \frac{1}{4} (\frac{1}{1\times 3} + \frac{1}{3\times 5} + \frac{1}{5\times 7} + ... + \frac{1}{(2k-1)(2k+1)}) - \frac{k}{4} \frac{1}{(2k+1)(2k+3)} $

$\displaystyle \frac{1}{8} (\frac{1}{1} - \frac{1}{3} + \frac{1}{3} -\frac{1}{5}+...+ \frac{1}{(2k-1)}- \frac{1}{(2k+1)}) - \frac{k}{4} \frac{1}{(2k+1)(2k+3)} $

$\displaystyle \frac{1}{8} (\frac{1}{1} - \frac{1}{(2k+1)}) - \frac{k}{4} \frac{1}{(2k+1)(2k+3)} $

$\displaystyle \frac{k}{4(2k+1)} - \frac{k}{4(2k+1)(2k+3)} $

$\displaystyle \frac{k(k+1)}{2(2k+1)(2k+3)} $

[gon]

$\frac{n}{(2n-1)(2n+1)(2n+3)} = \frac{1}{2}[\frac{(2n-1)+1}{(2n-1)(2n+1)(2n+3)}]$

$= \frac{1}{2}[\frac{1}{(2n+1)(2n+3)} + \frac{1}{(2n-1)(2n+1)(2n+3)}]$

$=\frac{1}{2}[\frac{1}{2}(\frac{1}{2n+1}-\frac{1}{2n+3}) + \frac{1}{4}(\frac{1}{(2n-1)(2n+1)} - \frac{1}{(2n+1)(2n+3)})]$

ดังนั้น $S_{\infty} = \frac{1}{2}[\frac{1}{2}(\frac{1}{3}) + \frac{1}{4}(\frac{1}{3})]$

[Sabre]

ขอบคุณทุกท่านครับ