Partial fraction decomposition

Jan 2010
142
0
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
Jun 2008
16,216
6,764
North Texas
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?
 
Jan 2010
142
0
\(\displaystyle \frac{1}{n^2+1}\) 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
Jun 2008
16,216
6,764
North Texas
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 ...

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

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
That is what it says in the problem...
the problem statement is wrong.

this sum is true ...

\(\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1\)
 
Jan 2010
142
0
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
Jun 2008
16,216
6,764
North Texas
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 \(\displaystyle n = 0\) ... \(\displaystyle A = 1\)

\(\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\)
 
Jan 2010
142
0
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
Jun 2008
16,216
6,764
North Texas
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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