Since $n!$ represents $$1\cdot2\cdot3\cdots n,$$ I am wondering if there is a way to represent $$1+2+3+\dots+n?$$ What are some usual notations for the computation of some common sequences? Any other examples?

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    $\begingroup$ $\sum_{i=1}^n i$. Also, $n(n+1)/2$. Depends on what you want to do with it. $\endgroup$ Apr 4 '14 at 7:25
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    $\begingroup$ Your sum is the Triangle Number $T_n$. $\endgroup$ Apr 4 '14 at 7:30
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    $\begingroup$ Only a comment, but I swear I've seen (maybe on another math.se post?) the notation $n?$ for this sum as well. $\endgroup$ Apr 4 '14 at 17:03
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    $\begingroup$ @StevenStadnicki According to Donald Knuth in Section 1.2.5 of Volume 1 of The Art of Computer Programming, the notation $$n? = \dfrac{1}{2}n(n+1)$$ is known as the termial function of $n$. However, I personally prefer the Triangular number interpretation for $T_n$. $\endgroup$
    – Xoque55
    Apr 4 '14 at 21:47
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    $\begingroup$ possible duplicate of What is the term for a factorial type operation, but with summation instead of products?; also Factorial, but with addition $\endgroup$
    – MJD
    Apr 6 '14 at 5:51

$T_n$, where the letter T stands for Triangular.

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    $\begingroup$ Good answer! However, when using that notation, it is good if you always specify that by $T_n$, you mean the $n$-th triangular number (unlike with $n!$, when it is perfectly clear what you mean). $\endgroup$
    – 5xum
    Apr 4 '14 at 7:43
  • $\begingroup$ Thanks, I didn't know this one before I checked your reference to wiki! $\endgroup$ Apr 4 '14 at 7:57

One way to write it would be simply by using the sumation notation, meaning $$1+2+3+\dots+n=\sum_{k=1}^n k.$$ Of course, that is equivalent to writing the factorials with the product notation, meaning $$n!=\prod_{k=1}^n k,$$ so I don't think that is what you were asking.

If you already know that $1+2+\dots+n=\frac{n(n+1)}{2}$, then you can just write $\frac{n(n+1)}{2}$ instead of the sum. The fact that the sum can be expressed as this rather short fraction is in my oppinion the real reason why a shortened notation does not exist. Unlike in the case of $n!$, which cannot be expressed by a polynomial in $n$, this one can be, thus shortening the notation is not needed.

EDIT: I was of course proven wrong by Chris Culter in his answer. However, I would like to add that the notation $T_n$, where $T$ stands for triangular, is not as common as $n!$. Any mathematitian in the world will know that if you write $5!$, you really mean $5\cdot 4\cdot 3\cdot 2\cdot 1$, however, if you say $T_n$, that notation can be used for other things, such as Chebyshev polynomials.

  • $\begingroup$ Thanks. It makes my mind clear $\endgroup$ Apr 4 '14 at 7:58
  • $\begingroup$ whortening $\mapsto$ shortening $\:$ ? $\;\;\;\;$ $\endgroup$
    – user57159
    Apr 5 '14 at 3:08

$$\sum_{n=1}^{k} n = 1+2+3+...+ k$$

Also, $$k! = \prod\limits_{n=1}^k n$$

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    $\begingroup$ Right, which is why using $\sum$ is not in the spirit of the OP's question. The evaluation $\frac{1}{2}n(n+1)$ is closer in spirit. $\endgroup$ Apr 4 '14 at 7:36

Another way to write it would be ${n+1 \choose 2}$:

${n \choose 0}=1$ for all $n$, and ${n \choose k+1} = \sum_{k=0}^{n-1}{n \choose k}$ (with ${0 \choose k} = 0$ for $k>0$). Therefore ${n\choose 1}=n$ and thus ${n+1 \choose 2}=\sum_{k=0}^n k$. Note that this actually is $0+1+\dots+n$, but obviously the additional term $0$ doesn't change the sum.


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