# Homework Help: Double Series

1. Nov 10, 2005

### 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

Last edited: Nov 10, 2005
2. Nov 10, 2005

### saltydog

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.

3. Nov 12, 2005

### 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:

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

4. Nov 12, 2005

### saltydog

Thanks Benorin. That's very nice.