Partial fraction decomposition

**Anemori** · May 9th 2010, 11:09 AM

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

like this?

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

**skeeter** · May 9th 2010, 11:12 AM

Originally Posted by Anemori

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

like this?

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

$\frac{1}{n^2+1}$ is not factorable ...

why do you need to perform the partial fraction decomposition?

**Anemori** · May 9th 2010, 11:29 AM

Originally Posted by skeeter

$\frac{1}{n^2+1}$ is not factorable ...

why do you need to perform the partial fraction decomposition?

I have this problem:

Consider $a_n=\frac{1}{n^2+1}$

Then,
a.) Rewrite $a_n$ using partial decomposition.

b.) Using a.) to help show the $Sum_\infty=1$

**skeeter** · May 9th 2010, 11:35 AM

Originally Posted by Anemori

I have this problem:

Consider $a_n=\frac{1}{n^2+1}$

Then,
a.) Rewrite $a_n$ using partial decomposition.

b.) Using a.) to help show the $Sum_\infty=1$

you need to check the problem statement again ...

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

**Anemori** · May 9th 2010, 12:00 PM

Originally Posted by skeeter

you need to check the problem statement again ...

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

That is what it says in the problem...

**skeeter** · May 9th 2010, 12:15 PM

Originally Posted by Anemori

That is what it says in the problem...

the problem statement is wrong.

this sum is true ...

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

**Anemori** · May 9th 2010, 12:28 PM

Originally Posted by skeeter

the problem statement is wrong.

this sum is true ...

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

I see thanks... did you use telescoping series?

still having problem rewriting it to partial fraction decomposition...

**skeeter** · May 9th 2010, 01:08 PM

Originally Posted by Anemori

I see thanks... did you use telescoping series?

still having problem rewriting it to partial fraction decomposition...

yes

$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

$1 = A(n+1) + Bn$

let $n = -1$ ... $B = -1$

let $n = 0$ ... $A = 1$

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

**Anemori** · May 9th 2010, 03:35 PM

Originally Posted by skeeter

yes

$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$

$1 = A(n+1) + Bn$

let $n = -1$ ... $B = -1$

let $n = 0$ ... $A = 1$

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

I'm sorry i got wrong equation. but it should be the same... its $a_n=\frac{1}{(n^2+n)}$

i look it up and it was the same... thanks!

**skeeter** · May 9th 2010, 03:45 PM

Originally Posted by Anemori

I'm sorry i got wrong equation. but it should be the same... its $a_n=\frac{1}{(n^2+n)}$

i look it up and it was the same... thanks!

... and you are very welcome.

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