I am dealing with a series that has a term: $$\sum\limits_{n=1}^\infty\sin\left(\frac{n\pi}{2}\right)$$
How would I express this without the sine function?
I know it alternates $1, 0, -1, 0, 1, \dots$
Can I express this as $-1$ raised to some power?
Asked
Active
Viewed 513 times
3
-
Not if you only want something of the form $(-1)^k$ (since you can't get 0). – Dennis Meng Oct 09 '13 at 01:32
2 Answers
4
I believe (with $n$ starting at 0) $$ \frac{i^n (1 + (-1)^n)}{2} $$ does the trick. If you want the index to begin at $1$ then simply replace $n$ with $n-1$ above.
tylerc0816
- 4,123
1
The series $\sum_{n=1}^{\infty} \sin\left(\frac{n \pi}{2}\right)$ does not converge. If you're dealing with something like $\sum_{n=1}^{\infty} \sin\left(\frac{n \pi}{2}\right) f(n)$ with $f$ some function, then you could replace the series by $\sum_{i=1}^{\infty}(-1)^{n-1}f(2n-1)$, since all other terms in the series are $0$ anyway.
ScroogeMcDuck
- 921