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I have always been confused with the arrangement of the summation indices after I interchange the order of the sums. For instance, I am encountering the following double summations that I wish to commute (The summation coefficients aren't important for my question so I avoid them, I am also assuming convergence.)

$$ \sum^{\infty}_{k=0}\sum^{2k}_{\nu=0}=\sum^{?}_{\nu=0}\sum^{?}_{k=0} $$

When I have nested sums like in the following double summations:

$$ \sum^{\infty}_{k=0}\sum^{k}_{\nu=0}=\sum^{\infty}_{\nu=0}\sum^{\infty}_{k=0}, $$

it is clear to me what is going on because we have the inequalities:

$$ 0\leq k \leq \infty, \hspace{5mm} 0 \leq \nu \leq k \hspace{5mm} \Rightarrow \hspace{5mm} 0 \leq \nu \leq k \leq \infty, $$

from which we can conclude that the bounds for the summation in $\nu$ will be $0 \leq \nu \leq \infty$. However in my example I have the bounds

$$ 0\leq k \leq\infty \\ 0 \leq \nu \leq 2k $$

which I cannot write in a concatenated single expression of inequalities. I think that the limits after I switch the summations are: $\sum^{\infty}_{\nu=0}\sum^{\infty}_{k=0}$ because the $k$ index runs until infinity so $\nu$ has to run to infinity as well. Would someone please help me to clarify whether or not these are the bounds?

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    $\begingroup$ Note that $\sum^{\infty}_{k=0}\sum^{k}_{\nu=0}=\sum^{\infty}_{\nu=0}\sum^{\infty}_{k=0}$ is actually incorrect; it should be $\sum^{\infty}_{k=0}\sum^{k}_{\nu=0}=\sum^{\infty}_{\nu=0}\sum^{\infty}_{k=v}$. $\endgroup$
    – Greg Martin
    Commented Dec 12, 2022 at 19:28
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    $\begingroup$ In general: for the outer (first, leftmost) sum you want to ask the question "what are all the possible values in the universe taken by the first index?"; for the second (inner, rightmost) sum you want to ask the question "given a particular value of the first index, what are all the values that the second index can take at the same time the first index takes its particular value?". I think this strategy is more robust than trying to combine inequalities together. $\endgroup$
    – Greg Martin
    Commented Dec 12, 2022 at 19:30
  • $\begingroup$ I would add that one way to think about such sums is to think of the summation as summing over a triangle. Taking the summation over this triangle in different directions gives you different reindexing. I use this picture to find the reindexing when in trouble. $\endgroup$
    – Zakhurf
    Commented Dec 12, 2022 at 22:11

1 Answer 1

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$\sum_{k=0}^\infty\sum_{\nu=0}^k$ runs over the set of pairs $(k,\nu)$ of integers satisfying $0\leqslant\nu\leqslant k$. For any fixed $\nu\geqslant 0$, a pair $(k,\nu)$ is in this set if and only if $k\geqslant\nu$. Hence $\sum_{k=0}^\infty\sum_{\nu=0}^k=\sum_{\nu=0}^\infty\sum_{k=\nu}^\infty$ (as commented).

Similarly, $\sum_{k=0}^\infty\sum_{\nu=0}^{2k}$ is over the set of $(k,\nu)$ with $0\leqslant\nu\leqslant 2k$. So, for a fixed $\nu\geqslant 0$, a pair $(k,\nu)$ is in this set if and only if $2k\geqslant\nu$, that is, $k\geqslant\lceil\nu/2\rceil$. Thus $\color{blue}{\sum_{k=0}^\infty\sum_{\nu=0}^{2k}=\sum_{\nu=0}^\infty\sum_{k=\lceil\nu/2\rceil}^\infty}$.

Sometimes (when the inequalities are confusing) it's worth plotting the set the sum runs over.

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