15 \(\blacklozenge\) Advanced Techniques
Double Sums (Pringsheim, Pg 471) CAUCHY DOUBLE SUMMATION FORMULA
15.1 Summation by Parts
Summation by parts
Abels Summation Lemma
Applicaiton: integer power sums (423)
Summation by Curves (pg 473 in Amazing)
15.2 Double Sums
Another useful application of dominated convergence is to switching the order of a double sum. A double sequence is a map \(\NN\times\NN\to \RR\), where we write \(a_{m,n}\) for the value \(a(m,n)\). Such sequences like \(n/(n+m)\) occured in our original example about switching limits above.
Given a double sequence, one may want to define an double sum
\[\sum_{m,n\geq 0}a_{m,n}\]
But, how should one do this? Because we have two indices, there are two possible orders we could attempt to compute this sum:
\[\sum_{n\geq 0}\sum_{m\geq 0}a_{m,n} \hspace{0.5cm}\textrm{or}\hspace{0.5cm}\sum_{m\geq 0}\sum_{n\geq 0}a_{m,n}\]
Definition 15.1 (Double Sum) Given a double sequence \(a_{m,n}\) its double sum \(\sum_{m,n\geq 0}a_{m,n}\) is defined if both orders of iterated summation converge, and are equal. In this case, the value of the double sum is defined to be their common value:
\[\sum_{m,n\geq 0}a_{m,n}:=\sum_{n\geq 0}\sum_{m\geq 0}a_{m,n} =\sum_{m\geq 0}\sum_{n\geq 0}a_{m,n}\]
We should be worried from previous experience that in general these two things need not be equal, so the double sum may not exist! Indeed, we can make this worry precise, by seeing that to relate one to the other is really an exchange of order of limits:
\[\sum_{m\geq 0}=\lim_M \sum_{0\leq m\leq M}\hspace{1cm}\sum_{n\geq 0}=\lim_N \sum_{0\leq n\leq N}\]
And so, expanding the above with these definitions (and using the limit laws to pull a limit out of a finite sum) we see
\[\sum_{n\geq 0}\sum_{m\geq 0}a_{m,n}=\lim_N \sum_{0\leq n\leq N}\left(\lim_M \sum_{0\leq m\leq M}a_{m,n}\right)\] \[=\lim_N\lim_M\left( \sum_{0\leq n\leq N}\sum_{0\leq m\leq M} a_{m,n}\right)=\lim_N\lim_M \sum_{\begin{smallmatrix}0\leq m\leq M\\ 0\leq n\leq N\end{smallmatrix}}a_{m,n}\]
Where in the final line we have put both indices under a single sum to indicate that it is a finite sum, and the order does not matter. Doing the same with the other order yields the exact same finite sum, but with the order of limits reversed:
\[\sum_{m\geq 0}\sum_{n\geq 0}a_{m,n}=\lim_M\lim_N \sum_{\begin{smallmatrix}0\leq m\leq M\\ 0\leq n\leq N\end{smallmatrix}}a_{m,n}\]
Because this is an exchange-of-limits-problem, we can hope to provide conditions under which it is allowed using Tannery’s theorem.
Theorem 15.1 Let \(a_{m,n}\) be a double sequence, and assume that either \[\sum_{m\geq 0}\sum_{n\geq 0}|a_{m,n}|\hspace{1cm}\textrm{or}\hspace{1cm}\sum_{n\geq 0}\sum_{m\geq 0}|a_{m,n}|\] converges. Then the double sum also converges \[\sum_{m,n\geq 0}a_{m,n}\] (meaning either both orders of iterated sum converge, and are equal)
Exercise 15.1 (Cauchy’s Double Summation Formula) Use Dominated Convergence to prove the double summation formula (Theorem 15.1): without loss of generality, assume that \(\sum_{m\geq 0}\sum_{n\geq 0}|a_{m,n}|\) converges, and use this to show that both orders of iterated sum converge and are equal \[\sum_{m\geq 0}\sum_{n\geq 0}a_{m,n}=\sum_{n\geq 0}\sum_{m\geq 0}a_{m,n}\]
Hint: Assuming \(\sum_{m\geq 0}\sum_{n\geq 0}|a_{m,n}|\) converges, set \(M_m=\sum_{n\geq 0}|a_{m,n}|\) and show the various hypotheses of Dominated convergence apply
Exercise 15.2 (Applying the Double Sum) Since switching the order of limits involves commuting terms that are arbitrarily far apart, techniques like double summation allow one to prove many identities that are rather difficult to show directly. We will make a crucial use of this soon, in understanding exponential functions. But here is a first example:
For any \(k\in\NN\), prove the following equality of infinite sums:
\[\frac{z^{1+k}}{1-z}+\frac{(z^2)^{1+k}}{1-z^2}+\frac{(z^3)^{1+k}}{1-z^3}+\cdots =\frac{z^{1+k}}{1-z^{1+k}}+\frac{z^{2+k}}{z^{2+k}}+\frac{z^{3+k}}{1-z^{3+k}}+\cdots\]
Hint: first write each side as a summation: \[\sum_{n\geq 1}\frac{z^{n(k+1)}}{1-z^n}=\sum_{m\geq 1}\frac{z^{m+k}}{1-z^{m+k}}\]
*Then setting \(a_{m,n}=z^{n(m+k)}\), show that Cauchy summation applies to the double sum \(\sum_{m,n}\geq 0 a_{m,n}\) and compute the sum in each order, arriving that the claimed equality.