Double Series and Fubini

Sum the grid by rows first or by columns first — with absolute convergence the answer is the same. ∑m,n1m2n2=(π26)2=π436\sum_{m,n} \frac{1}{m^2n^2} = \left(\frac{\pi^2}{6}\right)^2 = \frac{\pi^4}{36}. Conditional convergence lets the order change the answer.

∑m∑nam,n=∑n∑mam,n=sup⁡finite F∑Fam,n\sum_{m}\sum_{n} a_{m,n} = \sum_{n}\sum_{m} a_{m,n} = \sup_{\text{finite } F} \sum_{F} a_{m,n}
∑m,n≥11m2n2=π436≈2.7058\sum_{m,n \ge 1} \frac{1}{m^2n^2} = \frac{\pi^4}{36} \approx 2.7058

Examples & Exercises

Prove Fubini for infinite sums (nonnegative terms) completely: if am,n≥0a_{m,n} \ge 0 then ∑m∑nam,n=∑n∑mam,n\sum_m \sum_n a_{m,n} = \sum_n \sum_m a_{m,n}. Show both sides equal the supremum SS of sums over finite rectangles.

S=sup⁡M,N∑m≤M∑n≤Nam,n∈[0,∞]S = \sup_{M,N} \sum_{m \le M} \sum_{n \le N} a_{m,n} \in [0, \infty]
∑m=1∞∑n=1∞am,n=S=∑n=1∞∑m=1∞am,n\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n} = S = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n}
N=5N = 5: sum by rows ≈2.1422\approx 2.1422, sum by columns ≈2.1422\approx 2.1422 — both head to (π2/6)2=π4/36≈2.7058(\pi^2/6)^2 = \pi^4/36 \approx 2.7058.
Show proof
  1. Put T(M,N)=∑m≤M∑n≤Nam,nT(M,N) = \sum_{m \le M}\sum_{n \le N} a_{m,n} (a finite sum, ≥0\ge 0) and S=sup⁡M,NT(M,N)∈[0,∞]S = \sup_{M,N} T(M,N) \in [0,\infty]. TT is monotone in each of M,NM, N. For each fixed mm, the inner series ∑n=1∞am,n∈[0,∞]\sum_{n=1}^{\infty} a_{m,n} \in [0,\infty] converges (finitely or to +∞+\infty) since the terms are nonnegative.
  2. Put R=∑m=1∞∑n=1∞am,nR = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n}. For each M,NM, N: T(M,N)=∑m≤M∑n≤Nam,n≤∑m≤M∑n=1∞am,n≤RT(M,N) = \sum_{m \le M}\sum_{n \le N} a_{m,n} \le \sum_{m \le M}\sum_{n=1}^{\infty} a_{m,n} \le R (inner finite sum ≤\le inner infinite sum, outer finite ≤\le outer infinite — all terms nonnegative). Taking the supremum over M,NM, N gives S≤RS \le R.
  3. The other direction: for each MM, ∑m≤M∑n=1∞am,n=sup⁡NT(M,N)≤S\sum_{m \le M}\sum_{n=1}^{\infty} a_{m,n} = \sup_N T(M,N) \le S (the inner infinite sum is the supremum of the inner finite sums). Taking the supremum over MM gives R=sup⁡M∑m≤M∑n=1∞am,n≤SR = \sup_M \sum_{m \le M}\sum_{n=1}^{\infty} a_{m,n} \le S. Hence R=SR = S.
  4. Swapping the roles of rows and columns, the identical argument gives C=∑n=1∞∑m=1∞am,n=SC = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n} = S (the same supremum of the same T(M,N)T(M,N)). Hence R=C=SR = C = S — the two iterated sums agree. If S<∞S < \infty the common value is finite; if S=∞S = \infty both sides diverge to +∞+\infty. QED.

Extend to the absolutely convergent case (short complete proof): if ∑m,n∣am,n∣<∞\sum_{m,n} \lvert a_{m,n}\rvert < \infty, use the decomposition a=a+−a−a = a^+ - a^- to apply the first problem to each part and conclude ∑m∑nam,n=∑n∑mam,n\sum_m\sum_n a_{m,n} = \sum_n\sum_m a_{m,n}.

a+=max⁡(a,0),a−=max⁡(−a,0),∣a∣=a++a−,a=a+−a−a^+ = \max(a, 0), \quad a^- = \max(-a, 0), \quad \lvert a\rvert = a^+ + a^-, \quad a = a^+ - a^-
∑m∑na=∑m∑na+−∑m∑na−=∑n∑ma+−∑n∑ma−=∑n∑ma\sum_m\sum_n a = \sum_m\sum_n a^+ - \sum_m\sum_n a^- = \sum_n\sum_m a^+ - \sum_n\sum_m a^- = \sum_n\sum_m a
N=5N = 5: signs are present but ∑∣a∣\sum \lvert a\rvert converges, so sum by rows ≈0.7033\approx 0.7033 and sum by columns ≈0.7033\approx 0.7033 agree, both heading to π4/144≈0.6765\pi^4/144 \approx 0.6765.
Show proof
  1. Put am,n+=max⁡(am,n,0)a^+_{m,n} = \max(a_{m,n}, 0) and am,n−=max⁡(−am,n,0)a^-_{m,n} = \max(-a_{m,n}, 0). Then 0≤am,n±≤∣am,n∣0 \le a^{\pm}_{m,n} \le \lvert a_{m,n}\rvert, with ∣a∣=a++a−\lvert a\rvert = a^+ + a^- and a=a+−a−a = a^+ - a^-. By comparison, both ∑am,n+\sum a^+_{m,n} and ∑am,n−\sum a^-_{m,n} converge, bounded by ∑∣am,n∣<∞\sum \lvert a_{m,n}\rvert < \infty.
  2. Since a+a^+ and a−a^- are nonnegative, apply the first problem (Fubini, nonnegative terms) to each: ∑m∑na+=∑n∑ma+<∞\sum_m\sum_n a^+ = \sum_n\sum_m a^+ < \infty, and likewise for a−a^-. All iterated sums are finite, so subtraction is legitimate, and ∑m∑na=∑m∑na+−∑m∑na−=∑n∑ma+−∑n∑ma−=∑n∑ma\sum_m\sum_n a = \sum_m\sum_n a^+ - \sum_m\sum_n a^- = \sum_n\sum_m a^+ - \sum_n\sum_m a^- = \sum_n\sum_m a. QED.

Show the hypothesis is sharp: with an,n=1a_{n,n} = 1, an,n+1=−1a_{n,n+1} = -1 (n≥1n \ge 1) and 00 elsewhere, verify completely that every row sum is 00 (so ∑m∑nam,n=0\sum_m\sum_n a_{m,n} = 0) but column 11 sums to 11 and columns n≥2n \ge 2 sum to 00 (so ∑n∑mam,n=1\sum_n\sum_m a_{m,n} = 1), and note ∑∣a∣=∞\sum \lvert a\rvert = \infty.

an,n=1,an,n+1=−1 (n≥1),else 0a_{n,n} = 1, \quad a_{n,n+1} = -1 \ (n \ge 1), \quad \text{else } 0
∑m∑nam,n=∑m0=0≠1=1+∑n≥20=∑n∑mam,n\sum_m \sum_n a_{m,n} = \sum_m 0 = 0 \quad\ne\quad 1 = 1 + \sum_{n \ge 2} 0 = \sum_n \sum_m a_{m,n}
The alternating series in the picture above is absolutely convergent, so reordering is safe — but as the counterexample below shows, with ∑∣a∣=∞\sum\lvert a\rvert = \infty the order can change the answer.
Show proof
  1. Fix a row (mm fixed). The only nonzero entries are am,m=1a_{m,m} = 1 at n=mn = m and am,m+1=−1a_{m,m+1} = -1 at n=m+1n = m+1. Hence the row sum is ∑n=1∞am,n=1+(−1)=0\sum_{n=1}^{\infty} a_{m,n} = 1 + (-1) = 0 (a finite sum, so no convergence issue). This holds for every mm, so ∑m=1∞∑n=1∞am,n=∑m=1∞0=0\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n} = \sum_{m=1}^{\infty} 0 = 0.
  2. Fix a column (nn fixed). The nonzero entries: an,n=1a_{n,n} = 1 (row m=nm = n) and, for n≥2n \ge 2, an−1,n=−1a_{n-1,n} = -1 (row m=n−1m = n-1, from am,m+1=−1a_{m,m+1} = -1 with m=n−1m = n-1). Column 11: only a1,1=1a_{1,1} = 1, so the sum is 11. Column n≥2n \ge 2: 1+(−1)=01 + (-1) = 0. Hence ∑n=1∞∑m=1∞am,n=1+∑n=2∞0=1\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n} = 1 + \sum_{n=2}^{\infty} 0 = 1.
  3. The two iterated sums differ, 0≠10 \ne 1 — the order changes the answer. The broken hypothesis is absolute convergence: ∑m,n∣am,n∣=∑n=1∞(1+1)=∞\sum_{m,n} \lvert a_{m,n}\rvert = \sum_{n=1}^{\infty} (1 + 1) = \infty, so the second problem's hypothesis fails. Without the nonnegative (or absolutely convergent) hypothesis Fubini fails, and this example shows the hypothesis is sharp. QED.

6. Series