[Math] How to change the order of summation

summation

I have stumbled upon, multiple times, on cases where I need to change
the order of summation (usally of finite sums).

One problem I saw was simple
$$
\sum_{i=1}^{\infty}\sum_{j=i}^{\infty}f(i,j)=\sum_{j=1}^{\infty}\sum_{i=1}^{j}f(i,j)
$$

and I can go from the first sum to the second by noting that the
constraints are
$$
1\leq i\leq j<\infty
$$
so the first double sum does not constrain on $i$ and constrains
$j$ to $j\geq i$. The second double summation doesn't put any constrains
on $j$ but constrains $i$ relative to $j$ $(1\leq i\leq j)$.

While this approach works for simple examples such as this. I am having
problems using it where the bounds are more complicated.

The current problem interchanges the following
$$
\sum_{i=1}^{n-1}\sum_{k=2}^{n-i+1}\to\sum_{k=2}^{n}\sum_{i=1}^{n+1-k}
$$

I started by writing
$$
k\leq n-i+1
$$

and got
$$
i\leq n-k+1
$$

but all other bounds are not clear to me..

the problem is that I can't use this technique since I can't write
the inequalities in the same form of
$$
1\leq i\leq f(j)\leq n
$$

where $n$ is some bound (possibly $\infty$).

My question is how to approach the second example by a technique that
should be able to handle similar cases

Best Answer

Changing the order in the first double sum is manageable. We could therefore use it as some kind of prototype. We transform the second double sum, so that the index range is similar to the first one.

first double sum:

The following presentation of the index range might be helpful. \begin{align*} \sum_{i=1}^{\infty}\sum_{j=i}^{\infty}f(i,j)=\sum_{\color{blue}{1\leq i\leq j<\infty}}f(i,j)=\sum_{j=1}^{\infty}\sum_{i=1}^{j}f(i,j) \end{align*} If we focus on the middle double sum and look at the index range $1\leq i\leq j<\infty$ we observe the left-hand side as well as the right-hand side can be easily derived.

We do some rearrangements to derive a similar representation in the

second double sum:

We obtain \begin{align*} \sum_{i=1}^{n-1}\sum_{k=2}^{n-i+1}g(i,k)&=\sum_{i=1}^{n-1}\sum_{k=2}^{i+1}g(n-i,k)\tag{1}\\ &=\sum_{i=1}^{n-1}\sum_{k=1}^{i}g(n-i,k+1)\tag{2}\\ &=\sum_{1\leq k\leq i\leq n-1}g(n-i,k+1)\tag{3}\\ &=\sum_{k=1}^{n-1}\sum_{i=k}^{n-1}g(n-i,k+1)\tag{4} \end{align*}

Comment:

  • In (1) we change the order of the first sum $i\rightarrow n-i$. Note, that reversing the order this way \begin{align*} \sum_{i=1}^{n-1}a(i)&=a(1)+a(2)+\cdots+a(n-1)\\ &=a(n-1)+a(n-2)+\cdots+a(1)\\ &=\sum_{i=1}^{n-1}a(n-i) \end{align*} does not change the lower and upper index of $i$, but each occurrence of $i$ within the sum has to be substituted with $n-i$. So, we replace $a(i)$ with $a(n-i)$.

  • In (2) we shift the index $k$ by one, so that we also can start with $k=1$.

  • In (3) we write the double sum as we did in the first case.

  • In (4) it's easy to change the order of the double sum based upon the representation in (3).

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