Ytukyg

To determine convergence of
$$sum_{n=1}^infty a_n,$$
one can use the integral test if $f(n)=a_n$ satisfies certain properties.

Now, if I would like to determine convergence of double series
$$sum_{i=1}^infty sum_{j=1}^infty a_{ij},$$
do we have some 'integral test' for it?

I think one can evaluate
$$intint f(i,j),di,dj.$$

Let $f : [0,infty)times[0,infty) to [0,infty)$ be continuous, and satisfy
$$
x_1 < x_2 quad Longrightarrow quad f(x_1,y) ge f(x_2,y)
\
y_1 < y_2 quad Longrightarrow quad f(x,y_1) ge f(x,y_2).
$$
consider the two series
$$
sum_{i=0}^infty sum_{j=0}^infty f(i,j)
tag{1}$$
$$
sum_{i=1}^infty sum_{j=1}^infty f(i,j)
tag{$1'$}$$
and the integral
$$
int_0^infty int_0^infty f(x,y);dy;dx .
tag{2}$$

Note, because $f$ is nonnegative, (1), ($1'$) and (2) exist, but possibly equal $+infty$. So to say they "converge" means they have a finite value.

Now, given $i,j$, note that for all $x,y$ with $i le x le i+1, j le y le y+1$ we have
$$
f(i,j) ge f(x,y) ge f(i+1,j+1).
$$
therefore
$$
f(i,j) ge int_i^{i+1} int_j^{j+1} f(x,y);dy;dx ge f(i+1,j+1)
$$
Summing these, we get
$$
sum_{i=0}^inftysum_{j=0}^infty f(i,j) ge int_0^infty int_0^infty f(x,y);dy;dx
tag{3}$$
and
$$
int_0^infty int_0^infty f(x,y);dy;dx ge sum_{i=1}^inftysum_{j=1}^infty f(i,j)
tag{4}$$
So: we may use (3) to show: if (1) converges, then (2) converges. And we may use (4) to show: if (2) converges, then ($1'$) converges.

note

Those who like "simple" statments, may be disappointed. It is possible that (2) converges but (1) diverges because $sum_i f(i,0)$ and/or $sum_j f(0,j)$ diverge.

Define a function $f(x,y)$ such that $$f(i,j)=a_{ij}$$furthermore define $b_i=sum_{j=1}^{infty}f(i,j)$. This means that $$sum_{i=1}^{infty}b_i=sum_{i=1}^{infty}sum_{j=1}^{infty}f(i,j)$$is convergent iff $sum_{j=1}^{infty}f(x,j)$ is monotone decreasing (a sufficient condition is $f(x_1,y)ge f(x_2,y)$ when $x_1le x_2$) and $$int_{1}^{infty}sum_{j=1}^{infty}f(x,j)dx=sum_{j=1}^{infty}int_{1}^{infty}f(x,j)dx$$finally by defining $c_n=int_{1}^{infty}f(x,n)dx$, the summation $sum_{1}^{infty}c_n$ is bounded iff $$int_{1}^{infty}f(x,y)dx$$ is monotone decreasing over $y$ (with a similar sufficient condition) and $$int_{1}^{infty}int_{1}^{infty}f(x,y)dxdy$$is bounded.

If $f(t,x)$ is continuous in $D=[1,infty)times[1,infty)$. Also if $f(t,x)$ is non increasing in $D$ meaning that if $t_1,t_2,x_1,x_2in D$
$$
t_1<t_2Rightarrow f(t_1,x)geq f(t_2,x)textrm{, }forall x in D
$$
$$
x_1<x_2Rightarrow f(t,x_1)geq f(t,x_2)textrm{, }forall tin D
$$
If we assume also that
$$
f(n,m)rightarrow 0textrm{ as }n,mrightarrow inftytag 1
$$
and
$$
sum^{n}_{i=1}f(i,1)textrm{, }sum^{m}_{j=1}f(1,j)tag 2
$$
$$
sum^{n}_{i=1}f(i,m)textrm{, }sum^{m}_{j=1}f(n,j) tag 3
$$

are bounded for all both $n,m>>1$ positive integers, then exists constant$-C$ such that
$$
sum^{n,m}_{i,j=2}f(i,j)=int^{n}_{1}int^{m}_{1}f(t,x)dxdt+C+o(1)textrm{, as both }n,mrightarrowinfty.tag 4
$$
Proof.

Easily using the monotonicity of $f$ we find

$$
int^{i+1}_{i}int^{j+1}_{j}(f(t,x)-f(i+1,j+1))dxdtgeq0
$$
and
$$
int^{i+1}_{i}int^{j+1}_{j}(f(t,x)-f(i,j))dxdtleq0
$$
Hence
$$
sum^{n}_{i=2}sum^{m}_{j=2}f(i,j)leq int^{n}_{1}int^{m}_{1}f(t,x)dxdtleqsum^{n-1}_{i=1}sum^{m-1}_{j=1}f(i,j)
$$
Hence if we set
$$
k_{n,m}:=int^{n}_{1}int^{m}_{1}f(t,x)dxdt-sum^{n}_{i=2}sum^{m}_{j=2}f(i,j),
$$
then
$$
0leq k_{n,m}leq -f(1,1)+f(n,m)+sum^{n}_{i=1}f(i,1)+sum^{m}_{j=1}f(1,j)-sum^{n}_{i=1}f(i,m)-sum^{m}_{j=1}f(n,j).
$$
Also
$$
k_{n+1,m}-k_{n,m}=int^{n+1}_{1}int^{m}_{1}f(t,x)dxdt-sum^{n+1}_{i=2}sum^{m}_{j=2}f(i,j)-
$$
$$
-left(int^{n}_{1}int^{m}_{1}f(t,x)dxdt-sum^{n}_{i=2}sum^{m}_{j=2}f(i,j)right)=
$$
$$
=int^{n+1}_{n}int^{m}_{1}f(t,x)dxdt-sum^{m}_{j=2}f(n+1,j)=
$$
$$
=int^{n+1}_{n}left(sum^{m}_{j=2}int^{j}_{j-1}left(f(t,x)-f(n+1,j)right)dxright)dt=
$$
$$
=int^{n+1}_{n}left(sum^{m}_{j=2}int^{j}_{j-1}left| f(t,x)-f(n+1,j)right|dxright)dtgeq 0.
$$
Hence $k_{n+1,m}geq k_{n,m}$, for all $n,m$ positive integers. In the same way $k_{n,m+1}geq k_{n,m}$, for all $n,m$ positive integers.

Hence the double sequence $k_{n,m}$ is monotonic and bounded. Hence it must be convergent. From this we get $(4)$.