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I am trying to figure out what this sum means:

$$\sum_{1 \le x <y\le5} 1$$

Up to now, I have only seen sums with a start and endpoint (start and end value) for the summation. For example:

$$\sum_{n=k}^{\infty}n$$

What does it mean if I have $1 \le x <y\le5$ as a lower bound of my sum?

EDIT: I have been trying to understand the answer that gt6989b has given but I don't arrive at the same answer. I basically have a double sum to evaluate:

$$\sum_{x=1}^4\sum_{y=x+1}^51=\sum_{x=1}^4(1+1+1+1)=\sum_{x=1}^4 4=16?$$

I am pretty sure I am doing something wrong but I am stuck. Can anyone point out my mistake?

Nullspace
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  • You can read it as "The number of $(x,y)$ such that $1\leq x < y \leq 5$". (This is of course not a complete answer to your question but you already have one below). – Yanko Oct 18 '18 at 16:27

2 Answers2

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$$ \sum_{1 \le x < y \le 5} 1 = \sum_{x=1}^4 \sum_{y={x+1}}^5 1 = 10 $$

In other words, you always have $x < y$ and $x \ge 1, y \le 5$ is enforced for every valid pair $(x,y)$. Hence, the shorthand expression is $$1 \le x < y \le 5$$ which tightly collapses all constraints into one multiple inequality.

gt6989b
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  • Thank you for the answer. I have another question about the sums. I understand why the first sum goes from $x=1$ to $4$ (since $5> x \ge 1 $ but why does the second sum start from $y=x+1$? Why doesn't it start from $y=2$ (since $1<y \le 5$? Also, I am a bit confused how you arrived at $10$. Isn't the first sum $1+1+1+1$ and the second sum $1+1+1+1$ which would give me $4 \times 4 = 16$? – Nullspace Oct 18 '18 at 16:40
  • Or am I evaluating this some wrong? – Nullspace Oct 18 '18 at 16:52
  • @Nullspace the second start cannot start from $2$ because when $x=2$, the pair $(2,2)$ is not valid, since you must have $x<y$... – gt6989b Oct 18 '18 at 16:53
  • But aren't there only 4 pairs: $$(1,2);(2,3);(3,4);(4,5)?$$ – Nullspace Oct 20 '18 at 11:43
  • @Nullspace why is $(1,5)$ invalid? – gt6989b Oct 22 '18 at 03:04
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$\displaystyle\sum_{1 \le x <y\le5} 1 $ is the number of pairs $(x,y)$ such that $1 \le x <y\le5$.

It is implicitly assumed that $x,y$ are integers.

lhf
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  • But aren't there only 4 pairs: $$(1,2);(2,3);(3,4);(4,5)?$$ – Nullspace Oct 20 '18 at 11:42
  • @Nullspace, no, there is also $(1,3), (1,4), (1,5),(2,4), \dots$. – lhf Oct 20 '18 at 11:45
  • Thank you! Of course! I am so stupid... So the full list is: $$$$ \begin{array}{c|lcr} x & y \ \hline 1 & 2 \ 1 & 3 \ 1 & 4 \ 1 & 5 \ 2&3\ 2&4\ 2&5 \ 3&4 \ 3&5 \4 &5
    \end{array} $$$$ I am still unsure how to evaluate the sum though or am I just supposed to count the pairs like I did in the table?     – Nullspace Oct 20 '18 at 11:50