absolute convergence of double series

Let us consider the double series

${\displaystyle \sum _{i,j=1}^{\mathrm{\infty }}}{u}_{ij}$(1)

of real or complex numbers $u_{ij}$.  Denote the row series $u_{k1}+u_{k2}+\mathrm{\dots }$ by $R_k$, the column series $u_{1k}+u_{2k}+\mathrm{\dots }$ by $C_k$ and the diagonal series $u_{11}+u_{12}+u_{21}+u_{13}+u_{22}+u_{31}+\mathrm{\dots }$by DS.  Then one has the

Theorem.  All row series, all column series and the diagonal series converge absolutely and

$$\sum _{k=1}^{\mathrm{\infty }}{R}_k=\sum _{k=1}^{\mathrm{\infty }}{C}_k=\text{DS},$$

if one of the following conditions is true:

• The diagonal series converges absolutely.

• There exists a positive number $M$ such that every finite sum of the numbers $|u_{ij}|$ is $\leqq M$.

• The row series $R_k$ converge absolutely and the series $W_1+W_2+\mathrm{\dots }$ with

  $$\sum _{j=1}^{\mathrm{\infty }}|u_{kj}|=W_k$$

  is convergent.  An analogical condition may be formulated for the column series $C_k$.

Example.  Does the double series

$$\sum _{m=2}^{\mathrm{\infty }}\sum _{n=3}^{\mathrm{\infty }}{n}^{-m}$$

converge?  If yes, determine its sum.

The column series ${\displaystyle \sum _{m=2}^{\mathrm{\infty }}}{\left({\displaystyle \frac{1}{n}}\right)}^m$ have positive terms and are absolutely converging geometric series having the sum

$$\frac{{(1/n)}^2}{1-1/n}=\frac{1}{n(n-1)}=\frac{1}{n-1}-\frac{1}{n}=W_n.$$

The series $W_3+W_4+\mathrm{\dots }$ is convergent, since its partial sum is a telescoping sum

$$\sum _{n=3}^N{W}_n=\sum _{n=3}^N\left(\frac{1}{n-1}-\frac{1}{n}\right)=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\mathrm{\dots }+\left(\frac{1}{N-1}-\frac{1}{N}\right)$$

equalling simply $\frac{1}{2}-\frac{1}{N}$ and having the limit $\frac{1}{2}$ as  $N\to \mathrm{\infty }$.  Consequently, the given double series converges and its sum is $\frac{1}{2}$.