Summation from n=1 ti infinity of (sin(1))^n. How can I break this up? I think it will be a geometric series...
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$\displaystyle \left| r \right| < 1 \Rightarrow \quad \sum\limits_{n = 1}^\infty {r^n } = \frac{r}{{1 - r}}$
I thought it was r ^ (n-1)? And what is my r and my a?
Originally Posted by veronicak5678 Summation from n=1 ti infinity of (sin(1))^n. How can I break this up? I think it will be a geometric series... $\displaystyle |\sin(1)|=|r|<1\Rightarrow\sum)_{n=1}^{\infty}\sin ^n(1)=\frac{\sin(1)}{1-\sin(1)}$
Got it. Thank you!
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