กำหนด$ \sum_{n = 2}^{\infty} (\frac{1}{n^4-n^2} )= A$ จงหาค่าของ $\sum_{n = 2}^{\infty} ( \frac{1}{n^2}) $
$ก.\frac{3}{4} −A $
$ข.\frac{3}{4}+A$
$ค.\frac{1}{2}−A $
$ง.\frac{1}{2}+A$
โจทย์สวยเจงๆ ขอลองทำหน่อย
จาก $\frac{1}{n^4-n^2} = \frac{1}{n^2 (n^2-1)} = \frac{1}{n^2 -1} - \frac{1}{n^2}$
ดังนั้น $ \sum_{n = 2}^{\infty}( \frac{1}{n^4-n^2}) = \sum_{n = 2}^{\infty} (\frac{1}{n^2 -1} - \frac{1}{n^2})$
$A = \sum_{n = 2}^{\infty}( \frac{1}{n^2 -1}) - \sum_{n = 2}^{\infty}(\frac{1}{n^2})$
$\therefore \sum_{n = 2}^{\infty}(\frac{1}{n^2}) = \sum_{n = 2}^{\infty}( \frac{1}{n^2 -1}) - A$
พิจารณา $\sum_{n = 2}^{\infty} (\frac{1}{n^2 -1})$
$\sum_{n = 2}^{\infty}( \frac{1}{n^2 -1})$ = $\sum_{n = 2}^{\infty} (\frac{1}{2}[\frac{1}{n -1} - \frac{1}{n +1} ])$
$= \frac{1}{2}[(\frac{1}{1} -\not\frac{1}{3}) + (\frac{1}{2} -\not\frac{1}{4}) + (\not\frac{1}{3} -\not\frac{1}{5}) + ... ]) $
$= \frac{1}{2} [\frac{1}{1} + \frac{1}{2}] = \frac{1}{2}\cdot \frac{3}{2} = \frac{3}{4}$
$\therefore \sum_{n = 2}^{\infty} (\frac{1}{n^2}) = \frac{3}{4} - A $
