A Anemori Jan 2010 142 0 May 9, 2010 #1 I have this \(\displaystyle a_n=\frac{1}{n^2+1} \) and I have to rewrite this to partial fraction decomposition. like this? \(\displaystyle \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1} \)

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 May 9, 2010 #2 Anemori said: I have this \(\displaystyle a_n=\frac{1}{n^2+1} \) and I have to rewrite this to partial fraction decomposition. like this? \(\displaystyle \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1} \) Click to expand... \(\displaystyle \frac{1}{n^2+1}\) is not factorable ... why do you need to perform the partial fraction decomposition?

Anemori said: I have this \(\displaystyle a_n=\frac{1}{n^2+1} \) and I have to rewrite this to partial fraction decomposition. like this? \(\displaystyle \frac{n(x)}{d(x)}= \frac{1}{n}+\frac{1}{n+1} \) Click to expand... \(\displaystyle \frac{1}{n^2+1}\) is not factorable ... why do you need to perform the partial fraction decomposition?

A Anemori Jan 2010 142 0 May 9, 2010 #3 skeeter said: \(\displaystyle \frac{1}{n^2+1}\) is not factorable ... why do you need to perform the partial fraction decomposition? Click to expand... 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 said: \(\displaystyle \frac{1}{n^2+1}\) is not factorable ... why do you need to perform the partial fraction decomposition? Click to expand... 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 May 9, 2010 #4 Anemori said: 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 \(\displaystyle Sum_\infty=1\) Click to expand... you need to check the problem statement again ... \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1\)

Anemori said: 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 \(\displaystyle Sum_\infty=1\) Click to expand... you need to check the problem statement again ... \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1\)

A Anemori Jan 2010 142 0 May 9, 2010 #5 skeeter said: you need to check the problem statement again ... \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1\) Click to expand... That is what it says in the problem...

skeeter said: you need to check the problem statement again ... \(\displaystyle \sum_{n=1}^\infty \frac{1}{n^2+1} \ne 1\) Click to expand... That is what it says in the problem...

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 9, 2010 #6 Anemori said: That is what it says in the problem... Click to expand... the problem statement is wrong. this sum is true ... \(\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1\)

Anemori said: That is what it says in the problem... Click to expand... the problem statement is wrong. this sum is true ... \(\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1\)

A Anemori Jan 2010 142 0 May 9, 2010 #7 skeeter said: the problem statement is wrong. this sum is true ... \(\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1\) Click to expand... I see thanks... did you use telescoping series? still having problem rewriting it to partial fraction decomposition...

skeeter said: the problem statement is wrong. this sum is true ... \(\displaystyle \sum_1^{\infty} \frac{1}{n(n+1)} = 1\) Click to expand... 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 May 9, 2010 #8 Anemori said: I see thanks... did you use telescoping series? still having problem rewriting it to partial fraction decomposition... Click to expand... 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 said: I see thanks... did you use telescoping series? still having problem rewriting it to partial fraction decomposition... Click to expand... 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 \(\displaystyle n = 0\) ... \(\displaystyle A = 1\) \(\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\)

A Anemori Jan 2010 142 0 May 9, 2010 #9 skeeter said: 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 \(\displaystyle n = 0\) ... \(\displaystyle A = 1\) \(\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\) Click to expand... 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: May 9, 2010

skeeter said: 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 \(\displaystyle n = 0\) ... \(\displaystyle A = 1\) \(\displaystyle \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}\) Click to expand... 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!

skeeter MHF Helper Jun 2008 16,216 6,764 North Texas May 9, 2010 #10 Anemori said: 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! Click to expand... ... and you are very welcome. Reactions: Anemori

Anemori said: 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! Click to expand... ... and you are very welcome.