Suppose \[ a_{1,1}+a_{1,2}+\cdots,\qquad a_{2,1}+a_{2,2}+\cdots,\qquad\ldots \] is a countable collection of convergent series of nonnegative real numbers. Show that there is a convergent series \(x_1+x_2+\cdots\) of real numbers converging more slowly than any of the given series, in the sense that for every \(m\) one has \(x_n\ge a_{m,n}\) for all sufficiently large \(n\).
Hint: the problem is not affected by changing a finite number of terms of each given series.
Solution
1. For each \(m\geq1\), choose an integer \(N_m\) such that\[ \sum_{n=N_m}^{\infty}a_{m,n} \leq 2^{-m}. \]
The series \[ \sum_{n=1}^{\infty}a_{m,n} \] converges and has nonnegative terms, so its tails tend to \(0\). Hence a tail with sum at most \(2^{-m}\) exists.
2. Define\[ b_{m,n} \coloneqq \begin{cases} 0,&n<N_m,\\ a_{m,n},&n\geq N_m. \end{cases} \]Then\[ \sum_{n=1}^{\infty}b_{m,n} \leq 2^{-m}, \]and \(b_{m,n}=a_{m,n}\) for all sufficiently large \(n\).
1. Only the finitely many terms with \(n<N_m\) have been changed.
3. For every \(n\geq1\), the series\[ x_n \coloneqq \sum_{m=1}^{\infty}b_{m,n} \]converges to a finite nonnegative real number.
2. Therefore\[ 0 \leq \sum_{m=1}^{M}b_{m,n} \leq \sum_{m=1}^{M}2^{-m} < 1 \]for every \(M\). The increasing partial sums are bounded, so they converge.
4. The series\[ \sum_{n=1}^{\infty}x_n \]converges.
Fix \(N\geq1\). By step 3 and nonnegativity, \[ \begin{aligned} \sum_{n=1}^{N}x_n &= \lim_{M\to\infty} \sum_{n=1}^{N}\sum_{m=1}^{M}b_{m,n}\\ &= \lim_{M\to\infty} \sum_{m=1}^{M}\sum_{n=1}^{N}b_{m,n}\\ &\leq \lim_{M\to\infty} \sum_{m=1}^{M}2^{-m}\\ &\leq 1. \end{aligned} \] Thus the partial sums of the nonnegative series \(\sum_nx_n\) are bounded above by \(1\), so the series converges.
5. For every fixed \(m\geq1\),\[ x_n\geq a_{m,n} \]for all \(n\geq N_m\).
If \(n\geq N_m\), then step 2 gives \[ b_{m,n}=a_{m,n}. \] Since all \(b_{j,n}\) are nonnegative, \[ x_n = \sum_{j=1}^{\infty}b_{j,n} \geq b_{m,n} = a_{m,n}. \]
6. Therefore the convergent series\[ \boxed{x_1+x_2+\cdots} \]converges more slowly than every given series in the required sense.
4. For each \(m\), step 5 supplies the threshold \(N_m\) beyond which \(x_n\geq a_{m,n}\).
7. Q.E.D.
Step 6 is exactly the required conclusion.