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Find the sum:

$$\sum_\limits{i=0}^\infty\sum_\limits{j=0}^\infty \frac{1}{a^i\cdot a^j}$$ where $i \not= j$ and $a>1$

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    If the involved series are absolutely convergent, $$\sum_{\substack{i,j\geq 0\\ i\neq j}} f(i)\,f(j) = \left(\sum_{i\geq 0}f(i)\right)^2-\sum_{i\geq 0}f(i)^2.$$ – 2017-02-11

1 Answers 1

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$$\sum_\limits{i=0}^\infty\sum_\limits{j=0 \\ i\not=j}^\infty \frac{1}{a^i\cdot a^j}$$ $$\sum_\limits{i=0}^\infty\sum_\limits{j=0}^\infty \frac{1}{a^i\cdot a^j}-\sum_\limits{j=0}^\infty \frac{1}{a^{2j}}$$ $$=\left(\sum_\limits{i=0}^\infty \frac{1}{a^i}\right)\left(\sum_\limits{j=0}^\infty \frac{1}{a^j}\right)-\frac{1}{1-\frac{1}{a^2}}$$ $$=\frac{1}{1-\frac{1}{a}} \cdot \frac{1}{1-\frac{1}{a}}-\frac{a^2}{a^2-1}$$ $$=\left(\frac{a}{a-1}\right)^2-\frac{a^2}{a^2-1}$$ $$=\frac{a^2}{a-1}\left(\frac{1}{a-1}-\frac{1}{a+1}\right)$$ $$=\frac{2a^2}{(a-1)^2(a+1)}$$

Hope this helps you.

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    Thank you sir, I too found an answer to this in the exact way as yours but I was looking for an answer which used integration as I found it in an integration worksheet. – 2017-02-11
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    But this does not require integration. In any way whatsoever. – 2017-02-11
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    But sir my teacher told me it's wrong because you cannot split sigmas in multiplication. – 2017-02-11
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    Each summation has its own dummy index, one for $i$ and the other $j$. while summing over one, the other remains constant. Hence you can take that out and perform the sum just as I have done. The multiplication is not that the sum has been broken into multiplicands, rather the $2$ sums are independent, like $(a_1+a_2+a_3)(b_1+b_2+b_3)$. – 2017-02-11
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    Well, thanks for reminding that. My initial answer was a bit incomplete. Now I have added what was required. – 2017-02-11
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    Sir I would like to point out one more thing, can I split if the upper limits of both sigmas change to 1. – 2017-02-11
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53462/discussion-between-user328114-and-schrodingerscat). – 2017-02-11
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    sir please answer my question. – 2017-02-11
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    Yes you can split it. – 2017-02-11