- #1
jinksys
- 122
- 0
I understand that if given the following series:
\sum _{n=1}^{\infty } \frac{1}{n(n+3)}
I can break it up using partial fraction decomposition into:
\sum _{n=1}^{\infty } \frac{1}{3n}-\frac{1}{3n+9}
If I start listing the values of n_1,n_2, etc, I can see the cancellation pattern and find the sum.
This process seems tedious, is there another way of doing these problems?
Wikipedia says:
* Let k be a positive integer. Then
\sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k}
where Hk is the kth harmonic number. All of the terms after 1/(k − 1) cancel.
But I'm not sure what the 1/(k - 1) means, nor do I understand the harmonic number bit either.
\sum _{n=1}^{\infty } \frac{1}{n(n+3)}
I can break it up using partial fraction decomposition into:
\sum _{n=1}^{\infty } \frac{1}{3n}-\frac{1}{3n+9}
If I start listing the values of n_1,n_2, etc, I can see the cancellation pattern and find the sum.
This process seems tedious, is there another way of doing these problems?
Wikipedia says:
* Let k be a positive integer. Then
\sum^\infty_{n=1} {\frac{1}{n(n+k)}} = \frac{H_k}{k}
where Hk is the kth harmonic number. All of the terms after 1/(k − 1) cancel.
But I'm not sure what the 1/(k - 1) means, nor do I understand the harmonic number bit either.
