Expressing the summation $1-2-3-4-5..$ using strictly only one summation operator, in the simplest way possible.

Question

My friend got me on this question of writing the summation, using strictly one summation operator only. So;

$$\sum_{k=0}^n x = 1-2-3-4-5-...$$

So in here what will be the most simplest function or expression for $x$ suitable to get the result? Note that no

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

or similar expression is allowed as it has a part that is not under a summation operator. This has been a challenge to me honestly. While my work on this was this;

$$\sum_{k=0}^n \frac{\frac{3}{2}-k}{\sqrt{{(\frac{3}{2}-k)}^2}}k = 1-2-3-4-5-...$$

This also kinda looks not neat, I'm also trying to tone it down, so any suggestions?

Answer 1

What about $$\sum_{k=0}^n(-1)^{2^k+1}(k+1)?$$

Answer 2

The sequence of $1$ followed by $-1$ is A153881, from there you can construct your sum for example as \begin{align} S&=\sum_{k = 1}^n (1 - 2\cdot \text{sgn}(k-1))\cdot k \\ &=\sum_{k = 1}^n (-1)^{\text{sgn}(k-1)}\cdot k \\ &=\sum_{k = 1}^n \left(2\left\lfloor\frac{1}{k}\right\rfloor-1\right)\cdot k \\ &=\sum_{k = 1}^n (-1)^{p_k} k \end{align} where $\text{sgn}(x)$ is a Sign function and $p_n$ is $n$-th prime.

Answer 3

How about with the use of the Unit step function centered at 2: $$ \sum_{k=1}^n (-1)^{u_2(k)}\cdot k$$

Answer 4

A variation on $\left(2\lfloor\frac 1k\rfloor-1\right)k$ from Sil's.

This one can also be $(2\cdot0^k−1)$ upon the convention $0^0=1$. (or you can use $\delta_0(k)$ instead).

Note that $(-1)^{\delta_0(k)}$ offers also the same kind of alternative.

We could also put everything inside the floor function: $$\sum\limits_{k=0}^n\left\lfloor\frac{1-3k-2k^2}{2k+1}\right\rfloor$$