Determining the sigma notation for the series: 1 + 2 - 3 + 4 - 5 + 6 - 7 + 8 - 9 . . . . .?

Question

As stated above, the series is:

1 + 2 - 3 + 4 - 5 + 6 - 7 + 8 - 9 + 10 - 11 + 12 - 13 + 14 - 15 . . . ± (n - 1) ∓ n

What would the sigma notation be for this series, starting at 1, ending at n?

Thank you very much for your time!

score 2 · Answer 1 · 2012-10-03 03:22:50Z

up vote 2 down vote

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

edited Oct 3 '12 at 3:22

answered Oct 3 '12 at 3:17

user39572

	I don't understand the downvote, either. If the first term is +1, one can just add 2 to this. – Ross Millikan Oct 3 '12 at 3:21
	Nah. Meant it to start with the +1. – Jessica Hope Oct 3 '12 at 3:22
	@JessicaHope Adjusted. – user39572 Oct 3 '12 at 3:23
	Thank you very much, Julien! – Jessica Hope Oct 3 '12 at 3:28

Théophile · Answer 2 · 2012-10-03 03:29:53Z

up vote 1 down vote

A more abstruse solution: $$\sum_{k=1}^n (-1)^{k \cdot \textrm{sgn} (k-1)}k$$ where $\textrm {sgn}$ is the signum function.

answered Oct 3 '12 at 3:29

Théophile
1,767311

Avatar · Answer 3 · 2012-10-03 03:22:57Z

up vote 0 down vote

i think $$\sum_{k=1}^{n}(-1)^{k+\lfloor\frac{1}{k}\rfloor}\cdot k$$ works

edited Oct 3 '12 at 3:22

answered Oct 3 '12 at 3:17

Avatar
8,3451836

3 Answers