Let us consider the double series

(1)

Theorem. All row series, all column series and the diagonal series converge absolutely and $$ \sum_{k=1}^\infty R_k \;=\; \sum_{k=1}^\infty C_k \;=\; \mbox{DS}, $$ 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 $W_1\!+\!W_2\!+\ldots$ with terms $$ \sum_{j=1}^\infty|u_{kj}| = W_k $$ is convergent. An analogical condition may be formulated for the column series $C_k$ .

Example. Does the double series $$ \sum_{m=2}^\infty\sum_{n=3}^\infty n^{-m} $$ converge? If yes, determine its sum.

The column series $\displaystyle\sum_{m=2}^\infty\left(\frac{1}{n}\right)^m$ 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 $W_3\!+\!W_4\!+\ldots$ is convergent, since its partial sum is a telescoping sum $$ \sum_{n=3}^N W_n \;=\; \sum_{n=3}^N\left(\frac{1}{n\!-\!1}-\frac{1}{n}\right) \;=\; \left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right) +\ldots+\left(\frac{1}{N\!-\!1}-\frac{1}{N}\right) $$ equalling simply $\frac{1}{2}\!-\!\frac{1}{N}$ and having the limit $\frac{1}{2}$ as $N \to \infty$ . Consequently, the given double series converges and its sum is $\frac{1}{2}$ .