Exercise 8.2

Question

Exercise 2: 
    Let $a_{ij}$ be the number in the $i$th row and $j$th column of the array
    $$\begin{matrix}
        -1 & \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \cdots \[0.3em]
        \phantom{-}\frac{1}{2} & -1 & \phantom{-}0 &\phantom{-} 0 & \cdots \[0.3em]
        \phantom{-}\frac{1}{4} & \phantom{-}\frac{1}{2} & -1 & \phantom{-}0 & \cdots \[0.3em]
        \phantom{-}\frac{1}{8} & \phantom{-}\frac{1}{4} & \phantom{-}\frac{1}{2} & -1 & \cdots \[0.3em]
        \phantom{-}\vdots & \phantom{-}\vdots & \phantom{-}\vdots & \phantom{-}\vdots & \ddots
    \end{matrix}$$
    so that $$a_{ij}=
    \begin{cases}
        0 & (i<j), \
        -1 & (i=j), \
        2^{j-i} & (i>j).
    \end{cases}$$
    Prove that $$\sum_i\sum_ja_{ij}=-2,\quad\quad\sum_j\sum_ia_{ij}=0.$$

ghostofgarborg · Accepted Answer

(	extit{Matt ``Frito'' Lundy}) \
    Along any row, the positive entries form a finite geometric sequence, and the sum of all the entries along a row is
    $$\sum_{j = 1}^\infty a_{ij} 
    = -\left( \frac{1}{2}ight) ^{i-1},$$
    and so we have
    $$\sum_{i = 1}^\infty \sum_{j = 1}^\infty a_{ij}
    = -\sum_{i = 1}^\infty \left( \frac{1}{2}ight) ^{i-1}
    = -2.$$
    Whereas along any column, the positive entries form an infinite geometric sequence, and the sum of all the entries along a column is
    $$\sum_{i = 1}^{\infty} a_{ij} = -1 + \sum_{n=1}^\infty \left( \frac{1}{2}ight) ^n 
    = -1 + \left( \frac{1}{1 - \frac{1}{2}} - 1ight) = 0,$$
    so that
    $$\sum_{j = 1}^\infty \sum_{i = 1}^{\infty} a_{ij} = 0.$$
    Note that this double sequence $\{a_{ij}\}$ fails to meet the criteria of theorem 8.3 because if
    $$\sum_{j = 1}^\infty \left| a_{ij} ight| = b_i$$
    then $\sum b_i$ diverges.

Exercise 8.2

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Exercise 8.2

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