If for all natural numbers j and k, aj,k is a non-negative real number and aj,k ≤ aj+1,k, then[2]:168

lim j → ∞ ∑ k a j , k = ∑ k lim j → ∞ a j , k . {\displaystyle \lim _{j\to \infty }\sum _{k}a_{j,k}=\sum _{k}\lim _{j\to \infty }a_{j,k}.}

The theorem states that if you have an infinite matrix of non-negative real numbers such that

the COLUMNS are weakly increasing and bounded, and

for each row, the series whose terms are given by this row has a convergent sum,

then the limit of the sums of the rows is equal to the sum of the series whose term k is given by the limit of column k (which is also its SUPREMUM). The series has a convergent sum if and only if the (weakly increasing) sequence of row sums is bounded and THEREFORE convergent.

As an example, consider the infinite series of rows

( 1 + 1 N ) n = ∑ k = 0 n ( n k ) / n k = ∑ k = 0 n 1 k ! × n n × n − 1 n × ⋯ × n − k + 1 n , {\displaystyle \left(1+{\frac {1}{n}}\right)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}/n^{k}=\sum _{k=0}^{n}{\frac {1}{k!}}\times {\frac {n}{n}}\times {\frac {n-1}{n}}\times \cdots \times {\frac {n-k+1}{n}},}

where n approaches infinity (the limit of this series is e). Here the matrix entry in row n and column k is

( n k ) / n k = 1 k ! × n n × n − 1 n × ⋯ × n − k + 1 n ; {\displaystyle {\binom {n}{k}}/n^{k}={\frac {1}{k!}}\times {\frac {n}{n}}\times {\frac {n-1}{n}}\times \cdots \times {\frac {n-k+1}{n}};}

the columns (fixed k) are indeed weakly increasing with n and bounded (by 1/k!), while the rows only have finitely many nonzero terms, so condition 2 is satisfied; the theorem now says that you can compute the limit of the row sums

( 1 + 1 / n ) n {\displaystyle (1+1/n)^{n}}

by taking the sum of the column limits, namely

1 k ! {\displaystyle {\frac {1}{k!}}}

.