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I would like to know what the equation is for as series of infinite terms which are multiplied by the order of the terms: $$ \sum_{i=0}^{\infty} \sum_{j=0}^{\infty}(ij) a^ib^j $$ $a$ and $b$ are both fractions. Thanks to the answers provided on the question " Simple approximation to a series of infinite terms ", I assume that the this simplifies to: $$ \sum_{i=0}^{\infty} ia^i \cdot \sum_{j=0}^{\infty}jb^j $$ A simple formula similar to the answers provided in the previous question would be much appreciated.

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Jacques MALAPRADE
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    http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn –  Jul 28 '14 at 14:45
  • Thnx. Following the link above I found the a similar answer here http://math.stackexchange.com/questions/67364/sequence-sum-question . – Jacques MALAPRADE Jul 28 '14 at 15:17

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Assuming that $a$ and $b$ are constants with an absolute value less than 1.

Looking at each summation individually we know that from the Neumann series

$\displaystyle \sum_{i = 0}^{\infty} a^i = \dfrac{1}{1-a} $

Assuming that the derivative of the above series can be portrayed as

$\displaystyle f'(a) = \sum_{i = 0}^{\infty} ia^{i-1} = \dfrac{1}{(1-a)^2} $

After multiplying by $a$ on each side we get

$af'(a) = \displaystyle \sum_{i = 0}^{\infty} ia^i = \dfrac{a}{(1-a)^2}$

We can do the same with

$bf'(b) = \displaystyle \sum_{j = 0}^{\infty} jb^j = \dfrac{b}{(1-b)^2}$

Thus

$\displaystyle \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} (ij)a^ib^j = \dfrac{ab}{(1-a)^2(1-b)^2}$

Jacques MALAPRADE
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