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I have the following nested sum that is mentioned in a statistic textbook without any justification. The transformation really doesnt seem obvious to me. Could someone maybe elaborate how this holds?

$$ C \sum_{k=2}^{\infty} \sum_{j=1}^{k-1}\left(\frac{1}{2}\right)^{k}=C \sum_{j=1}^{\infty} \sum_{k=j+1}^{\infty}\left(\frac{1}{2}\right)^{k}$$

Blue
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Phil
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    The summands are non-negative numbers, you can carry the sum in any order your want. On both sides, the index runs over same set of $(j,k) \in \mathbb{Z}^{2}$. Namely, those with $j < k$. As a result, the sum evaluates to same value. – achille hui Jul 20 '21 at 14:02
  • Related: https://math.stackexchange.com/questions/89814/positivity-and-interchange-of-summation. – Martin R Jul 20 '21 at 14:13

2 Answers2

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Draw a two dimensional plot $(k,j)$ and start to put dots where the sum is evaluated. You'll see that if you go in one direction, say fix $k$ and bound $j$ by $1..k-1$. But if you "rotate" your point of view, and fix $j$, you see that $j$ goes from 1 to $\infty$ but $k$ starts at the point $j+1$ and goes to $\infty$.

Tom Davis
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    The important point is that all numbers are non-negative (or that the double series converges absolutely). Otherwise changing the order of summation might very well change the sum. – Martin R Jul 20 '21 at 14:06
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Note that if $j \le k-1$, then we have $k \ge j+1$:

\begin{align} \{(k, j) \in \mathbb{Z}^2: 2 \le k, 1 \le j \le k-1\} = \{(k, j) \in \mathbb{Z}^2: 1 \le j , j+1 \le k \} \end{align}

More details:

  • If $2 \le k , 1 \le j \le k-1$, then we have $1 \le j$ and $(j \le k-1 \iff k \ge j+1)$.

  • If $1 \le j $ and $k \ge j+1$, then clearly we have $k \ge j+1 \ge 1+1=2$ and $1 \le j$ and $(k \ge j+1 \iff j \le k-1)$.

Try to learn to sketch the region and describe the region in two different orders. Also note that all the terms are nonnegative.

Siong Thye Goh
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