The sum of the convergent series [tex]\sum_{n=1}^{\infty}~(sin(1))^n[/tex] is 5.31
For given question,
We have been given a series [tex]\sum_{n=1}^{\infty}~(sin(1))^n[/tex]
[tex]\sum_{n=1}^{\infty}~(sin(1))^n=sin(1)+(sin(1))^2+...+(sin(1))^n[/tex]
We need to find the sum of given convergent series.
Given series is a geometric series with ratio r = sin(1)
The first term of the given geometric series is [tex]a_1=sin(1)[/tex]
So, the sum is,
= [tex]\frac{a_1}{1-r}[/tex]
= sin(1) / [1 - sin(1)]
This means, the series converges to sin(1) / [1 - sin(1)]
[tex]\sum_{n=1}^{\infty}~(sin(1))^n[/tex]
= [tex]\frac{sin(1)}{1-sin(1)}[/tex]
= [tex]\frac{0.8415}{1-0.8415}[/tex]
= [tex]\frac{0.8415}{0.1585}[/tex]
= 5.31
Therefore, the sum of the convergent series [tex]\sum_{n=1}^{\infty}~(sin(1))^n[/tex] is 5.31
Learn more about the convergent series here:
https://brainly.com/question/15415793
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