Calculate Double Series: \sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^(j+i)

[AnthonyS]

Hello, all--
I have the following series that I'm supposed to calculate:

\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^(j+i)

Could someone show me how it should be done?

Note: It's supposed to be p^(j+i) or p raised to j+i; however, I can't get the formatting to work.


Anthony

 

[saltydog]

Hello, all--
I have the following series that I'm supposed to calculate:
\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^(j+i)
Could someone show me how it should be done?
Note: It's supposed to be p^(j+i) or p raised to j+i; however, I can't get the formatting to work.
Anthony

Here's the formatting:

\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^{(j+i)}

Suppose p was between 0 and 1, say 1/2 and we start with i=1. The first member of the set would then be:

\sum_{j=1}^{\infty}\left(\frac{1}{2}\right)^{1+j}

second one, well you know what it it. Third one . . . They converge don't they? Suppose they all converge to numbers less than 1. I don't know but suppose they did and you were able to express the general form of those numbers say:

s_1=1/2

s_2=1/4

s_3=1/8

s_n=\left(\frac{1}{2}\right)^n

Then we could sum them right? Not saying that's what they would be but this is just one way I can think of.

 

[benorin]

The double seires you have presented is convergent only for \left| p\right| <1 (p may be real or complex), for these values of p we have:

\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^{j+i} = \sum_{i=1}^{\infty} \left( p^{i} \sum_{j=i}^{\infty} p^{j}\right) = \sum_{i=1}^{\infty} \left( p^{2i} \sum_{j=i}^{\infty} p^{j-i}\right)
=\sum_{i=1}^{\infty} \left( p^{2i} \sum_{j=0}^{\infty} p^{j}\right) = \left( \sum_{i=1}^{\infty} p^{2i}\right) \left( \sum_{j=0}^{\infty} p^{j}\right)
= \left(p^{2} \sum_{i=0}^{\infty} \left( p^{2}\right)^{i}\right) \left( \frac{1}{1-p}\right) = \left(\frac{p^{2}}{1-p^{2}} \right) \left( \frac{1}{1-p}\right) = \frac{p^{2}}{(1-p)^{2}(1+p)}

 

[saltydog]

The double seires you have presented is convergent only for \left| p\right| <1 (p may be real or complex), for these values of p we have:
\sum_{i=1}^{\infty} \sum_{j=i}^{\infty} p^{j+i} = \sum_{i=1}^{\infty} \left( p^{i} \sum_{j=i}^{\infty} p^{j}\right) = \sum_{i=1}^{\infty} \left( p^{2i} \sum_{j=i}^{\infty} p^{j-i}\right)
=\sum_{i=1}^{\infty} \left( p^{2i} \sum_{j=0}^{\infty} p^{j}\right) = \left( \sum_{i=1}^{\infty} p^{2i}\right) \left( \sum_{j=0}^{\infty} p^{j}\right)
= \left(p^{2} \sum_{i=0}^{\infty} \left( p^{2}\right)^{i}\right) \left( \frac{1}{1-p}\right) = \left(\frac{p^{2}}{1-p^{2}} \right) \left( \frac{1}{1-p}\right) = \frac{p^{2}}{(1-p)^{2}(1+p)}

Thanks Benorin. That's very nice.