question 6 of 13\ndetermine convergence or divergence of $sum_{n = 1}^{infty}sin(\frac{1}{n})$ using any…

question 6 of 13\ndetermine convergence or divergence of $sum_{n = 1}^{infty}sin(\frac{1}{n})$ using any method.\nthe series converges.\nthe series diverges.

question 6 of 13\ndetermine convergence or divergence of $sum_{n = 1}^{infty}sin(\frac{1}{n})$ using any method.\nthe series converges.\nthe series diverges.

Answer

Explanation:

Step1: Use limit - comparison test

We know that for small (x), (\sin x\sim x) as (x\to0). Let (a_n = \sin(\frac{1}{n})) and (b_n=\frac{1}{n}). Then (\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=\lim_{n\rightarrow\infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}}). Let (x = \frac{1}{n}), as (n\rightarrow\infty), (x\rightarrow0). And (\lim_{x\rightarrow0}\frac{\sin x}{x}=1).

Step2: Analyze the series of (b_n)

The series (\sum_{n = 1}^{\infty}b_n=\sum_{n=1}^{\infty}\frac{1}{n}) is the harmonic series, which is a well - known divergent series ((p) - series with (p = 1), and for (p) - series (\sum_{n=1}^{\infty}\frac{1}{n^p}), it diverges when (p\leqslant1)).

Step3: Apply the limit - comparison test result

Since (\lim_{n\rightarrow\infty}\frac{\sin(\frac{1}{n})}{\frac{1}{n}} = 1>0) and (\sum_{n = 1}^{\infty}\frac{1}{n}) diverges, by the limit - comparison test, the series (\sum_{n=1}^{\infty}\sin(\frac{1}{n})) diverges.

Answer:

The series diverges.