1. ## Partial fraction decomposition

I have this $\displaystyle a_n=\frac{1}{n^2+1}$ and I have to rewrite this to partial fraction decomposition.

like this?

$\displaystyle \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1}$

2. 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?

3. 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 $\displaystyle Sum_\infty=1$

4. 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 ...

$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1$

5. 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...

6. Originally Posted by Anemori
That is what it says in the problem...
the problem statement is wrong.

this sum is true ...

$\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1$

7. 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...

8. 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 $\displaystyle n = 0$ ... $\displaystyle A = 1$

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

9. 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!

10. 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.