Question

Determine whether the series is convergent or divergent: \(\sum\limits_{n = 1}^\infty {\sin \frac{1}{n}} \).

Short answer

The series \(\sum\limits_{n = 1}^\infty {\sin \frac{1}{n}} \)is divergent.

Step-by-step solution

Using Limit Comparison Test:

We have \({a_n} = \sin \frac{1}{n}\)

Let \({b_n} = \frac{1}{n}\)

Then using limit comparison test

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \frac{1}{n}}}{{\frac{1}{n}}}\)

Using Limit result\(\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1\):

Let \(\frac{1}{n} = \theta \) so \(\theta \to 0\) as \(n \to \infty \)

Then \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1\)

Comparing \({a_n}\)and \({b_n}\):

Now \(\mathop {\lim }\limits_{n \to \infty } \frac{{{a_n}}}{{{b_n}}} = 1 > 0\)

Since, \(\sum\limits_{n = 1}^\infty {{b_n}} = \sum\limits_{n = 1}^\infty {\frac{1}{n}} \) is harmonic series which is divergent.

So, the series \(\sum\limits_{n = 1}^\infty {\sin \frac{1}{n}} \) is also divergent.

Hence, the series is divergent.