hey guys i'm having a little trouble working out a question. the question is to find a formula for the nth partial sum of the series sum of 1/((4n+1)(4n-3)) between n=1 and infinity. any help would be great thanks
Is...
$\displaystyle \displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (\frac{1}{4n-3} - \frac{1}{4n+1})$ (1)
... so that...
$\displaystyle \displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{4}\ (1 - \frac{1}{4N+1})$ (2)
... and finally...
$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
probably just typo it's
$\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}$ (3)
Edit:
$\displaystyle \displaystyle \frac {1}{(4n+1)(4n-3)} = \frac {A}{4n+1} + \frac {B}{4n-3} $
$\displaystyle 4A+4B =0 \Rightarrow A=-B$
$\displaystyle \displaystyle -3A+B=1 \Rightarrow -4A=1 \Rightarrow A=-\frac {1}{4} \Rightarrow B = \frac {1}{4} $
...... ........