# Partial fraction decomposition

#### Anemori

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

like this?

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

#### skeeter

MHF Helper
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?

#### Anemori

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

#### skeeter

MHF Helper
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 ...

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

#### Anemori

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...

#### skeeter

MHF Helper
That is what it says in the problem...
the problem statement is wrong.

this sum is true ...

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

#### Anemori

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...

#### skeeter

MHF Helper
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 $$\displaystyle n = 0$$ ... $$\displaystyle A = 1$$

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

#### Anemori

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:

#### skeeter

MHF Helper
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.

• Anemori