1. ## partial sum question

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

2. Originally Posted by newcastle
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
first use partial fractions to write that sum than write few of them (depending how many you need to realize which is formula for nth partial sum)

show what have you done

$\displaystyle s = \sum _{n=1} ^{N} \frac {N}{4N+1}$

3. Is...

$\displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (\frac{1}{4n-3} - \frac{1}{4n+1})$ (1)

... so that...

$\displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{4}\ (1 - \frac{1}{4N+1})$ (2)

... and finally...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}$ (3)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by chisigma
Is...

$\displaystyle \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}\ (\frac{1}{4n-3} - \frac{1}{4n+1})$ (1)

... so that...

$\displaystyle \sum_{n=1}^{N} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}\ (1 -\frac{1}{5} + \frac{1}{5} -\frac{1}{9} + \frac{1}{9} -... - \frac{1}{4N+1}) = \frac{1}{7}\ (1 - \frac{1}{4N+1})$ (2)

... and finally...

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{7}$ (3)

Kind regards

$\chi$ $\sigma$

probably just typo it's

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{(4n+1)\ (4n-3)} = \frac{1}{4}$ (3)

Edit:

$\displaystyle \frac {1}{(4n+1)(4n-3)} = \frac {A}{4n+1} + \frac {B}{4n-3}$

$4A+4B =0 \Rightarrow A=-B$

$\displaystyle -3A+B=1 \Rightarrow -4A=1 \Rightarrow A=-\frac {1}{4} \Rightarrow B = \frac {1}{4}$

...... ........