1. ## Summation to infinity

Determine $\displaystyle \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3}$ given that $\displaystyle \frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}$

2. Originally Posted by Punch
Determine $\displaystyle \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3}$ given that $\displaystyle \frac{2}{4r^2-8r+3}=\frac{1}{2r-3}-\frac{1}{2r-1}$
With regard to your previous thread you know that, $\displaystyle \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)$ So take the limit as n tends to infinity.

3. Originally Posted by Sudharaka
With regard to your previous thread you know that, $\displaystyle \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)$ So take the limit as n tends to infinity.
as n tends to infinity, $\displaystyle \frac{1}{2n-1}$ tends to $\displaystyle 0.$

$\displaystyle \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3}=-\frac{1}{2}$

but ans is $\displaystyle -\frac{1}{6}$

4. Originally Posted by Punch
as n tends to infinity, $\displaystyle \frac{1}{2n-1}$ tends to $\displaystyle 0.$

$\displaystyle \sum_{n = 0}^\infty \frac{1}{4r^2-8r+3}=-\frac{1}{2}$

but ans is $\displaystyle -\frac{1}{6}$
In your previous thread we have found that, $\displaystyle \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)$

Here we have to find, $\displaystyle \sum_{r = 0}^{n}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{n}\frac{1}{4r^2-8r+3}$

Therefore taking the limit as n tend to infinity,

$\displaystyle \sum_{r = 0}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$

Hope I did not confuse you with my last post, there I did not notice that the summation must be taken from r=0.

5. Originally Posted by Sudharaka
In your previous thread we have found that, $\displaystyle \sum_{r=1}^{n}(\frac{1}{4r^2-8r+3})=\frac{1}{2}\left(-1-\frac{1}{2n-1}\right)$

Here we have to find, $\displaystyle \sum_{r = 0}^{n}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{n}\frac{1}{4r^2-8r+3}$

Therefore taking the limit as n tend to infinity,

$\displaystyle \sum_{r = 0}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}+\sum_{r = 1}^{\infty}\frac{1}{4r^2-8r+3}=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$

Hope I did not confuse you with my last post, there I did not notice that the summation must be taken from r=0.
thanks i didnt notice that it starts from zero too