# Thread: Finding Particular P-series

1. ## Finding Particular P-series

Hello,

So, I have this problem that is wanting me to find two particular p-series that satisfy some particular conditions. I don't really know how to approach this problem and I'm not sure I even understand what is being asked.

The problem:

"Find two p-series: $\sum_{k=1}^{\infty}c_k$ and $\sum_{k=1}^{\infty}d_k$ such that $\sum_{k=1}^{\infty}\frac{8(-1)^kk-7(k)^\frac{1}{2}}{k^\frac{3}{2}} = \sum_{k=1}^{\infty}((-1)^kc_k + d_k)$.

Your answer should be the formulas for $c_k$ and $d_k$ separated by commas.Both terms should be of the form $\frac{c}{k^p}$ for some constants $c$ and $p$. Note that $c_k$ should be first."

If anyone has any hints as to how to approach this (for example, is there a simple matter of algebraic finagling to be done or some other "hidden" method?), I would be grateful.

Thanks

2. Have you tried $c_k = \frac{8}{{k^{\frac{1}{2}} }}\,\& \,d_k = \frac{{ - 7}}{k}$?

3. Originally Posted by Plato
Have you tried $c_k = \frac{8}{{k^{\frac{1}{2}} }}\,\& \,d_k = \frac{{ - 7}}{k}$?

Oh, no. Even though those seem obvious, I hadn't tried any particular p-series, because I wasn't sure what the problem was asking. But, it appears those are the correct answers.

4. Originally Posted by auslmar
Oh, no. Even though those seem obvious, I hadn't tried any particular p-series, because I wasn't sure what the problem was asking. But, it appears those are the correct answers.
how did Plato see there solutions?

$\sum_{k=1}^{\infty}\frac{8(-1)^kk-7(k)^\frac{1}{2}}{k^\frac{3}{2}} = \sum_{k = 1}^\infty \bigg( \frac {8(-1)^kk}{k^{\frac 32}} - \frac {7k^{\frac 12}}{k^{\frac 32}} \bigg) = \sum_{k = 1}^\infty \bigg( (-1)^k {\color{red}\frac {8k}{k^{\frac 32}}} - {\color{red}\frac {7k^{\frac 12}}{k^{\frac 32}}} \bigg)$.

now do you see where they came from?

5. Originally Posted by Jhevon
how did Plato see there solutions?

$\sum_{k=1}^{\infty}\frac{8(-1)^kk-7(k)^\frac{1}{2}}{k^\frac{3}{2}} = \sum_{k = 1}^\infty \bigg( \frac {8(-1)^kk}{k^{\frac 32}} - \frac {7k^{\frac 12}}{k^{\frac 32}} \bigg) = \sum_{k = 1}^\infty \bigg( (-1)^k {\color{red}\frac {8k}{k^{\frac 32}}} - {\color{red}\frac {7k^{\frac 12}}{k^{\frac 32}}} \bigg)$.

now do you see where they came from?
Yeah, I can see now. It just wasn't occurring to me before.

6. Originally Posted by auslmar
those seem obvious, I hadn't tried any particular p-series, because I wasn't sure what the problem was asking. But, it appears those are the correct answers.
I have some real concerns about the series of problems you have posted.
It appears that whatever computer program is validating your answers has some real limitations. Therefore, I must say that I not sure what you are learning from this exercise. I hope that I am wrong.

7. Originally Posted by Plato
I have some real concerns about the series of problems you have posted.
It appears that whatever computer program is validating your answers has some real limitations. Therefore, I must say that I not sure what you are learning from this exercise. I hope that I am wrong.
Yes, I agree that the assignments and the system aren't designed as great exercises.