Summation of Summation: Calculating p^j*p^i

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SUMMARY

The discussion focuses on calculating the double summation of the series defined by the expression p^(j+i), where the outer summation runs from i=1 to infinity and the inner summation runs from j=i to infinity. The expression can be simplified by separating it into (p^j)*(p^i) and introducing a new variable k=j-i. This transformation allows for the summation of p^(2i+k) with k ranging from 0 to infinity and i from 1 to infinity, facilitating a clearer approach to the calculation.

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Calculate the summation of i=1 to inf of the summation of j=i to inf of p^(j+i).
Yes, it is the summation of a summation. p^(j+i) can be separated into (p^j)*(p^i).
 
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Write k=j-i, then you need to sum p2i+k with k=0 to ∞ and i=1 to ∞.
 

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