Solution to 8.29

(a) -> Suppose   is bounded by . Then so is   Since   is a bounded and increasing, it follows that it converges to some .

We show that the double sequence s(p,q) also converges to . Take Epsilon>0. Then there exists;N>0 such that &ForAll p>N, we have , i.e. . Take that N. Suppose p,q>N. Let us assume (no loss of generality) that p<=q. Then, by assumption, . But since , it follows that .

<- If the double sequence converges to (i.e. if the double series converges to ), then surely each partial sum is bounded by (since all terms are positive).

(b) Suppose converges to . Want to show that also converges.

Every partial sum  is in the interval : this is true since the absolute value of such partial sum is less than the sum od the absolute values of the terms, which in turn is less than .

If the set of all numbers is finite, then the set of all numbers is also finite, and so the same is true for the set of numbers . In that case all of these terms except finitely many should be equal to 0, for else would not converge. The rest of that case is easy.

Suppose now that the set of all numbers is infinite. Then (Bolzano), it must have an accumulation point in the interval . If we show that there are no two such, then we are done. Suppose otherwise: suppose and are two accumulation points of the double sequence {}. Denote their distance by . Then, for any (large) N, there are such that is less than units far from and such that   is less than units far from . So that we can find arbitrarily large such that and   are at least units apart. But the difference between   and   is certainly less than   (this is easy and involves inequalities with absolute values). On the other hand, since converges, could be made as small as we please by choosing  sufficiently large . In portraiture, it could certainly be made to be less that . So we get   for some . So, and must be equal.

(c) Since converges (earlier Calculus) it follows from Theorem 8.44 that the double series in that question also converges.