Proving Double Sum Equivalence

[James R]

Does anybody know how to prove the following?

\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(p,p-q)

f(n,m) is any function of n and m.

Note the change of the limiting values on the sums.

 

[NateTG]

I'd be very impressed if a suitable proof exists, considering that it's not true.

For example, if
f(x,y)=2^{y-2x}
then the LHS is clearly not convergent, but the RHS is.

 

[shmoe]

Does anybody know how to prove the following?

\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(p,p-q)

f(n,m) is any function of n and m.

Note the change of the limiting values on the sums.

That looks false to me, but it may be a little typo. Do you perhaps mean:

\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(q,p-q)

This kind of rearrangement usually follows from absolute convergence pretty easily, are you sure there's no restrictions on f(n,m)?

 

[NateTG]

That looks false to me, but it may be a little typo. Do you perhaps mean:

\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(q,p-q)

This kind of rearrangement usually follows from absolute convergence pretty easily, are you sure there's no restrictions on f(n,m)?

That's still false.
Consider, for example f(x,y) is 2^{-x!-y!} if y>x and 0 otherwise. Both sums are absolutely convergent, but the LHS is non-zero, and the RHS is zero.

If there is absolute convergence, then
\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(m,n)=\sum_{p=0}^\infty\sum_{q=0}^p f(p+q,p-q)+\sum_{p=0}^\infty\sum_{q=0}^p f(p+q+1,p-q)
works. Although, I'd be inclined to write that
\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(m,n)=\sum_{k=0}^{\infty}\sum_{m>0,n>0,m+n=k}f(m,n)

 

[shmoe]

That's still false.
Consider, for example f(x,y) is 2^{-x!-y!} if y>x and 0 otherwise. Both sums are absolutely convergent, but the LHS is non-zero, and the RHS is zero.

The RHS is not zero. Take the p=1 term, it's f(0,1)+f(1,0), one of these is non-zero. (Maybe you missed where I changed f(p,p-q) to f(q,p-q))

You can think of the double sum as a sum over the integer lattice (n,m) where n and m are >=0. The LHS sums along infinite vertical lines (e.g. (0,0), (0,1), (0,2),...) then sums the results. My version sums along the lines with slope -1 (the lines m+n=p), then sums the results. James version looks to sum everything under the line y=x say (though I'd expect it was a typo).

Your version

\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}f(m,n)=\sum_ {k=0}^{\infty}\sum_{m>0,n>0,m+n=k}f(m,n)

is the same as mine except you seem to be missing all points where one of n or m is zero. If you meant for n and m to be >= to zero instead of just >0, then it's the same as mine (take k=p, m=q)

 

[NateTG]

Yeah, those should have been greater than or equal to.

The RHS is not zero. Take the p=1 term, it's f(0,1)+f(1,0), one of these is non-zero. (Maybe you missed where I changed f(p,p-q) to f(q,p-q))

Yeah, I misread that as (q,q-p) which is no good.

 

[James R]

Damn! Sorry about that. It was a typo, and should read:

\sum\limits_{n=0}^{\infty} \sum\limits_{m=0}^{\infty} f(n,m) = \sum\limits_{p=0}^{\infty} \sum\limits_{q=0}^{p} f(q,p-q)

That is, it is f(q,p-q) rather than f(p,p-q).

So, given that, does anybody have a proof?

Here's one specific context, for example.

We know that exp(x+y) = exp(x)exp(y).

Writing exp(x+y) and exp(x) as Maclaurin series, I would like to prove that the left hand side equals the right hand side. If you try that, it leads to a double sum of the type given.

 

[shmoe]

You will be able to find proofs of this sort of thing under various assumptions in most intro analysis texts, see Rudin's Principles of Mathematical Analysis for example.