Proving that two double sums are equal

[DeadOriginal]

Homework Statement

Let [itex]\{a_{n,k}:n,k\in\mathbb{N}\}\subseteq[0,\infty)[/itex]. Prove that [itex]\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{n}a_{n,k}=\sum\limits_{k=0}^{\infty}\sum\limits_{n=k}^{\infty}a_{n,k}[/itex].

Homework Equations

The Attempt at a Solution

I am pretty certain that the claim is true because when I expand them out I get
[itex]\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{n}a_{n,k}=a_{0,0}+a_{1,0}+a_{1,1}+a_{2,0}+a_{2,1}+a_{2,2}+...[/itex] and
[itex]\sum\limits_{k=0}^{\infty}\sum\limits_{n=k}^{\infty}a_{n,k}=a_{0,0}+a_{1,0}+a_{2,0}+...+a_{1,1}+a_{2,1}+...[/itex]
which look to me like reorderings of each other. The problem is I am not sure about how I should approach proving that they are in fact equal.

 

[LCKurtz]

Homework Statement

Let [itex]\{a_{n,k}:n,k\in\mathbb{N}\}\subseteq[0,\infty)[/itex]. Prove that [itex]\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{n}a_{n,k}=\sum\limits_{k=0}^{\infty}\sum\limits_{n=k}^{\infty}a_{n,k}[/itex].

Homework Equations

The Attempt at a Solution

I am pretty certain that the claim is true because when I expand them out I get
[itex]\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{n}a_{n,k}=a_{0,0}+a_{1,0}+a_{1,1}+a_{2,0}+a_{2,1}+a_{2,2}+...[/itex] and
[itex]\sum\limits_{k=0}^{\infty}\sum\limits_{n=k}^{\infty}a_{n,k}=a_{0,0}+a_{1,0}+a_{2,0}+...+a_{1,1}+a_{2,1}+...[/itex]
which look to me like reorderings of each other. The problem is I am not sure about how I should approach proving that they are in fact equal.

Since you have infinite sums there, you are going to have to assume things converge. What you have is a problem of switching the order of summation. Draw a picture of the region in the n-k plane that the left sum describes then use the picture to reverse the order like you would do with double integrals.

Alternatively, you could define ##\epsilon_{nk} = 1## if ##k\le n## and ##0## if ##k>n##. Then you have$$
\sum_{n=0}^\infty \sum_{k=0}^n a_{nk} = \sum_{n=0}^\infty \sum_{k=0}^\infty a_{nk}\epsilon_{nk}$$Now reverse the order, which is easy, and rewrite the inner sum without the ##\epsilon_{nk}##'s by using the appropriate limits for ##n##.

 

[DeadOriginal]

Since you have infinite sums there, you are going to have to assume things converge. What you have is a problem of switching the order of summation. Draw a picture of the region in the n-k plane that the left sum describes then use the picture to reverse the order like you would do with double integrals.

Alternatively, you could define ##\epsilon_{nk} = 1## if ##k\le n## and ##0## if ##k>n##. Then you have$$
\sum_{n=0}^\infty \sum_{k=0}^n a_{nk} = \sum_{n=0}^\infty \sum_{k=0}^\infty a_{nk}\epsilon_{nk}$$Now reverse the order, which is easy, and rewrite the inner sum without the ##\epsilon_{nk}##'s by using the appropriate limits for ##n##.

I can see how the proof will go if reversing the order of summation works out but I am confused by that part. Does it necessarily follow that
$$\sum\limits_{n=0}^{\infty}\sum\limits_{k=0}^{\infty}a_{nk}\epsilon_{nk}
=\lim\limits_{l\rightarrow\infty}\sum\limits_{n=0}^{l}\left[\lim\limits_{m\rightarrow\infty}\sum\limits_{k=0}^{m}a_{nk}\epsilon_{nk}\right]
=\lim\limits_{l\rightarrow\infty}\lim\limits_{m\rightarrow\infty}\sum_{n=0}^{l}\sum\limits_{k=0}^{m}a_{nk}\epsilon_{nk}
=\lim\limits_{l\rightarrow\infty}\lim\limits_{m\rightarrow\infty}\sum_{k=0}^{m}\sum\limits_{n=0}^{l}a_{nk}\epsilon_{nk}?
$$

 

[LCKurtz]

There is a Fubini theorem for double sums which would justify formally switching the order. Here's one link:

http://www.google.com/url?sa=t&rct=...cbEOujI8KTGORVg&bvm=bv.45512109,d.cGE&cad=rja

 

[DeadOriginal]

Using Fubini's Theorem makes sense now. Thank you for the link.
Now if I were to prove it with your first hint does this look correct?
$$
\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}a_{nk}=a_{00}+a_{10}+a_{11}+a_{20}+a_{21}+a_{22}+...
$$
so it is equivalent to the summation of each element of the following matrix column by column.
$$
\begin{pmatrix}
a_{00} & a_{10} & a_{20} & a_{30} & \cdots \\
0 & a_{11} & a_{21} & a_{31} & \cdots \\
0 & 0 & a_{22} & a_{32} & \cdots \\
0 & 0 & 0 & a_{33} & \cdots \\
\vdots & \vdots & \vdots & \vdots \\
\end{pmatrix}
$$
To reverse the order we sum the elements of the following matrix row by row such that we get
$$
\begin{align*}
&a_{00}+a_{10}+a_{20}+\cdots+0+a_{11}+a_{21}+a_{31}+\cdots+0+0+a_{22}+a_{32}+\cdots+0+0+0+a_{33}+\cdots\\
&=a_{00}+a_{10}+a_{20}+\cdots+a_{11}+a_{21}+a_{31}+\cdots+a_{22}+a_{32}+\cdots+a_{33}+\cdots\\
&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}a_{nk}.
\end{align*}
$$
Hence
$$\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{nk}=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}a_{nk}.$$

 

[LCKurtz]

Since you have infinite sums there, you are going to have to assume things converge. What you have is a problem of switching the order of summation. Draw a picture of the region in the n-k plane that the left sum describes then use the picture to reverse the order like you would do with double integrals.

Alternatively, you could define ##\epsilon_{nk} = 1## if ##k\le n## and ##0## if ##k>n##. Then you have$$
\sum_{n=0}^\infty \sum_{k=0}^n a_{nk} = \sum_{n=0}^\infty \sum_{k=0}^\infty a_{nk}\epsilon_{nk}$$Now reverse the order, which is easy, and rewrite the inner sum without the ##\epsilon_{nk}##'s by using the appropriate limits for ##n##.

Using Fubini's Theorem makes sense now. Thank you for the link.
Now if I were to prove it with your first hint does this look correct?
$$
\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}a_{nk}=a_{00}+a_{10}+a_{11}+a_{20}+a_{21}+a_{22}+...
$$
so it is equivalent to the summation of each element of the following matrix column by column.
$$
\begin{pmatrix}
a_{00} & a_{10} & a_{20} & a_{30} & \cdots \\
0 & a_{11} & a_{21} & a_{31} & \cdots \\
0 & 0 & a_{22} & a_{32} & \cdots \\
0 & 0 & 0 & a_{33} & \cdots \\
\vdots & \vdots & \vdots & \vdots \\
\end{pmatrix}
$$
To reverse the order we sum the elements of the following matrix row by row such that we get
$$
\begin{align*}
&a_{00}+a_{10}+a_{20}+\cdots+0+a_{11}+a_{21}+a_{31}+\cdots+0+0+a_{22}+a_{32}+\cdots+0+0+0+a_{33}+\cdots\\
&=a_{00}+a_{10}+a_{20}+\cdots+a_{11}+a_{21}+a_{31}+\cdots+a_{22}+a_{32}+\cdots+a_{33}+\cdots\\
&=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}a_{nk}.
\end{align*}
$$
Hence
$$\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{nk}=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}a_{nk}.$$

I wouldn't argue it with symbolically written infinite matrices or sums full of dot dot dots. Also your matrix isn't written with the usual notation where the first subscript represents the row number. Altogether it gives me a headache trying to decipher your argument. Why don't you try my suggestion I have requoted above with the ##\epsilon_{nk}##'s. Just do what I suggest in the last line; it's easy.

 

[DeadOriginal]

Ok. Here is an attempt.
$$
\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{nk}\\
=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}a_{nk}\epsilon_{nk}=\sum_{k=0}^{\infty}\sum_{n=0}^{\infty}a_{nk}\epsilon_{nk} \text{ (by Fubini's Theorem)}\\
=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}a_{nk}$$
where [itex]\epsilon_{nk}=1[/itex] if [itex]k\leq n[/itex] and [itex]\epsilon_{nk}=0[/itex] if [itex]k>n[/itex].

 

[LCKurtz]

That's the idea. You explained that the second = is by Fubini's theorem. But your explanation should also include commentary as to exactly why the ##\epsilon_{nk}## make the first equality and last equality work. After all, you want your teacher to know that you really understand how the epsilons affect the indexing and aren't just shuffling symbols around.