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I am evaluating the following infinite sum but I keep getting an answer that is inconsistent with some other result that I know to be true. I would appreciate it if someone could check my solution. Here is the sum I want to evaluate.

$$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\theta^j\theta^i3^{i\wedge j}$$

$\theta \in (0,\frac{1}{3\sqrt{3}})$. $i\wedge j := \min(i,j)$.

$$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\theta^j\theta^i3^{i\wedge j} = \underbrace{\sum_{j=1}^{\infty}\sum_{i=1}^{j}\theta^j\theta^i3^{i\wedge j}}_{1} + \underbrace{\sum_{j=1}^{\infty}\sum_{i=j+1}^{\infty}\theta^j\theta^i3^{i\wedge j}}_{2}$$

$$\sum_{j=1}^{\infty}\sum_{i=1}^{j}\theta^j\theta^i3^{i\wedge j} = \sum_{j=1}^{\infty}\sum_{i=1}^{j}\theta^j\theta^i3^i = \sum_{j=1}^{\infty}\theta^j\frac{3\theta}{1-3\theta}(1-(3\theta)^j) = \frac{3\theta}{1-3\theta}\left(\frac{\theta}{1-\theta} - \frac{3\theta^2}{1-3\theta^2}\right) \tag{1}$$

$$\sum_{j=1}^{\infty}\sum_{i=j+1}^{\infty}\theta^j\theta^i3^{i\wedge j} = \sum_{j=1}^{\infty}\sum_{i=j+1}^{\infty}\theta^j\theta^i3^j = \sum_{j=1}^{\infty}(3\theta)^j\frac{\theta^{j+1}}{1-\theta} = \frac{\theta}{1-\theta} \frac{3\theta^2}{1-3\theta^2}\tag{2}$$

So then, $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\theta^j\theta^i3^{i\wedge j} = \frac{3\theta^2 + 3\theta^3}{(1-\theta)(1-3\theta^2)}$$

Is this correct?

Calculon
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    One way to check the answer is to consider terms of lowest order. For example, the term $(i,j)=(1,1)$ gives $3\theta^2$ (and this is the only such term on the LHS) but the RHS starts with $6\theta^2$. So the two don't match. – Semiclassical May 09 '16 at 21:11
  • @Semiclassical I have no idea where that $6$ came from. If it is $3$, which it should be, everything checks out. – Calculon May 09 '16 at 21:47

1 Answers1

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Here is an altenative derivation. The subset of terms with $i\wedge j=k$ contributes to the double sum as $$3^k\theta^{k+k}+\sum_{j=k}^{\infty}3^k\theta^{j+k}+\sum_{i=k}^{\infty}3^k\theta^{i+k}=(3\theta^2)^k\left(1+2\theta+2\theta^2+\cdots \right)=(3\theta^2)^k\left(\frac{1+\theta}{1-\theta}\right).$$ Summing over all positive $k$ then yields $\frac{3\theta^2}{1-3\theta^2}\left(\frac{1+\theta}{1-\theta}\right)$ in agreement with the above.

Semiclassical
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