I have this $\displaystyle a_n=\frac{1}{n^2+1} $ and I have to rewrite this to partial fraction decomposition. like this? $\displaystyle \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1} $
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Originally Posted by Anemori I have this $\displaystyle a_n=\frac{1}{n^2+1} $ and I have to rewrite this to partial fraction decomposition. like this? $\displaystyle \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1} $ $\displaystyle \frac{1}{n^2+1}$ is not factorable ... why do you need to perform the partial fraction decomposition?
Originally Posted by skeeter $\displaystyle \frac{1}{n^2+1}$ is not factorable ... why do you need to perform the partial fraction decomposition? I have this problem: Consider $\displaystyle a_n=\frac{1}{n^2+1}$ Then, a.) Rewrite $\displaystyle a_n $ using partial decomposition. b.) Using a.) to help show the $\displaystyle Sum_\infty=1$
Originally Posted by Anemori I have this problem: Consider $\displaystyle a_n=\frac{1}{n^2+1}$ Then, a.) Rewrite $\displaystyle a_n $ using partial decomposition. b.) Using a.) to help show the $\displaystyle Sum_\infty=1$ you need to check the problem statement again ... $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1$
Originally Posted by skeeter you need to check the problem statement again ... $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1$ That is what it says in the problem...
Originally Posted by Anemori That is what it says in the problem... the problem statement is wrong. this sum is true ... $\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1$
Originally Posted by skeeter the problem statement is wrong. this sum is true ... $\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1$ I see thanks... did you use telescoping series? still having problem rewriting it to partial fraction decomposition...
Originally Posted by Anemori I see thanks... did you use telescoping series? still having problem rewriting it to partial fraction decomposition... yes $\displaystyle \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$ $\displaystyle 1 = A(n+1) + Bn$ let $\displaystyle n = -1$ ... $\displaystyle B = -1$ let $\displaystyle n = 0$ ... $\displaystyle A = 1$ $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$
Originally Posted by skeeter yes $\displaystyle \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$ $\displaystyle 1 = A(n+1) + Bn$ let $\displaystyle n = -1$ ... $\displaystyle B = -1$ let $\displaystyle n = 0$ ... $\displaystyle A = 1$ $\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$ I'm sorry i got wrong equation. but it should be the same... its $\displaystyle a_n=\frac{1}{(n^2+n)} $ i look it up and it was the same... thanks!
Last edited by Anemori; May 9th 2010 at 03:44 PM. Reason: wrong equation
Originally Posted by Anemori I'm sorry i got wrong equation. but it should be the same... its $\displaystyle a_n=\frac{1}{(n^2+n)} $ i look it up and it was the same... thanks! ... and you are very welcome.
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