Question

determine whether the series is convergent or divergent. if it is convergent, find its sum. (if the quantity diverges, enter diverges.)

\\\sum_{k = 1}^{\infty} (\sin(200))^k\\

Explanation:

⚡ Using what you learned: Geometric Series

Step 1: Identify the series type and common ratio

The given series is:

$$ \sum_{k=1}^{\infty} (\sin(200))^k $$

This is a geometric series of the form \(\sum_{k=1}^{\infty} a r^{k-1}\) where:

• First term (\(k=1\)): \(a = \sin(200)\)
• Common ratio: \(r = \sin(200)\)

Step 2: Check for convergence

A geometric series converges if and only if \(|r| < 1\).

Since the range of the sine function is \([-1, 1]\), we have:

$$ -1 \le \sin(200) \le 1 $$

Because \(200\) (in radians) is not an odd multiple of \(\frac{\pi}{2}\), we know:

$$ |\sin(200)| < 1 $$

Thus, the series converges.

Step 3: Calculate the sum

The sum \(S\) of a convergent infinite geometric series starting at \(k=1\) is given by:

$$ S = \frac{a}{1 - r} $$

Substitute \(a = \sin(200)\) and \(r = \sin(200)\):

$$ S = \frac{\sin(200)}{1 - \sin(200)} $$

Answer:

$$ \frac{\sin(200)}{1 - \sin(200)} $$