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Let $a_{ij}$ be entries in a matrix in $i$th row and $j$th column such that $a_{ij} = \left\{ \begin{array}{l l} 0 & \quad \text{if $i$ < $j$}\\ -1 & \quad \text{if $i$ = $j$}\\ 2^{j-i} & \quad \text{if $j$ < $i$} \end{array} \right.$. Prove that $\sum \limits_{i} \sum \limits_{j} a_{ij} = -2$ and show $\sum \limits_{j} \sum \limits_{i} a_{ij} = 0$. If you have rudin handy, (baby rudin) its number 2 on page 196. Now intuitively I see this happening but I've never worked with doubles series (holding $i$ fixed while letting $j$ go to infinity and vis versa) and our professor didn't go over how to compute double series, and this is merely a computation problem.

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The matrix looks like this: $$\pmatrix{-1&0&0&0&\dots\cr1/2&-1&0&0&\dots\cr1/4&1/2&-1&0&\dots\cr1/8&1/4&1/2&-1&\dots\cr\vdots&\vdots&\vdots&\vdots&\vdots\cr}$$ Look at any column. Ignoring any leading zeros, the sum of the entries is $-1+(1/2)+(1/4)+(1/8)+\cdots=0$, since $(1/2)+(1/4)+(1/8)+\cdots=1$ as the sum of a geometric series. So this says that for every $j$, $\sum_ia_{ij}=0$ --- the sum just adds up all the numbers in column $j$. Then $\sum_j\sum_ia_{ij}=\sum_j0=0$.

Now, look at the rows. The terms in the first row sum to $-1$; the second row, $-1/2$; the third row, $-1/4$; the fourth row, $-1/8$; in general, the $i$th row, $-1/2^{i-1}$ (to be rogorous, you'd have to insert a proof by induction here, but I'm sure you can handle that). So, $\sum_i\sum_ja_{ij}=\sum_i(-1/2^{i-1})$, and that's a geometric series with sum $-2$. Done.

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