Question

Find the sum of the convergent series. (round your answer to four decimal places. ) [infinity] (sin(9))n n = 1

Answer 1

The sum of the convergent series $\sum_{n=1}^{\infty}~(sin(1))^n$ is 5.31

For given question,

We have been given a series $\sum_{n=1}^{\infty}~(sin(1))^n$

$\sum_{n=1}^{\infty}~(sin(1))^n=sin(1)+(sin(1))^2+...+(sin(1))^n$

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 $a_1=sin(1)$

So, the sum is,

= $\frac{a_1}{1-r}$

= sin(1) / [1 - sin(1)]

This means, the series converges to sin(1) / [1 - sin(1)]

$\sum_{n=1}^{\infty}~(sin(1))^n$

= $\frac{sin(1)}{1-sin(1)}$

= $\frac{0.8415}{1-0.8415}$

= $\frac{0.8415}{0.1585}$

= 5.31

Therefore, the sum of the convergent series $\sum_{n=1}^{\infty}~(sin(1))^n$ is 5.31

Learn more about the convergent series here:

brainly.com/question/15415793

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