Question

How do you use the limit comparison test to determine if $\sum \sin \left( {\dfrac{1}{n}} \right)$ from $[1,\infty )$ is convergent or divergent?

Answer 1

Hint : In order to solve this question we must know about the concept of the limit comparison test . It is a method of testing for the convergence of an infinite series . It Suppose that we have two series ${\sum _n}{a_n}$ and ${\sum _n}{b_n}$ with ${a_n} \geqslant 0,{b_n} > 0$ for all $n$ .
Then if $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}} = c$ with $0 < c < \infty ,$ , then either both series converge or both series diverge . And so we check for the convergence or divergence of $\sum \sin \left( {\dfrac{1}{n}} \right)$ from $[1,\infty )$ .

Complete step by step solution:
In the above question given , we are to determine if $\sum \sin \left( {\dfrac{1}{n}} \right)$ from $[1,\infty )$ is convergent or divergent using limit comparison test . We can see the comparison with conditions –
Suppose that we have two series $\sum {a_n}$ and $\sum {b_n}$ with ${a_n},{b_n} \geqslant 0$ , for all $n$ and ${a_n} \leqslant {b_n}$ for all $n$ . Then,
If $\sum {b_n}$ is convergent then so is $\sum {a_n}$ .
If $\sum {a_n}$ is divergent then so is $\sum {b_n}$ .
In other words, we have two series of positive terms and the terms of one of the series is always larger than the terms of the other series. Then if the larger series is convergent the smaller series must also be convergent. Likewise, if the smaller series is divergent then the larger series must also be divergent.
Let ${a_n} = \sin \left( {\dfrac{1}{n}} \right)$ and ${b_n} = \left( {\dfrac{1}{n}} \right)$ .
Then $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}} = \mathop {\lim }\limits_{n \to \infty } \dfrac{{\sin \left( {\dfrac{1}{n}} \right)}}{{\left( {\dfrac{1}{n}} \right)}}$ . There are multiple ways to approach this limit. The first is to replace the variables: as $n \to \infty ,\dfrac{1}{n} \to 0$ , so this can be rewritten as $\mathop {\lim }\limits_{a \to 0} \dfrac{{\sin \left( a \right)}}{{\left( a \right)}}$ which is a fundamental limit: $\mathop {\lim }\limits_{a \to 0} \dfrac{{\sin \left( a \right)}}{{\left( a \right)}} = 1$ So : $\mathop {\lim }\limits_{n \to \infty } \dfrac{{\sin \left( {\dfrac{1}{n}} \right)}}{{\left( {\dfrac{1}{n}} \right)}} = \mathop {\lim }\limits_{a \to 0} \dfrac{{\sin \left( a \right)}}{{\left( a \right)}} = 1$
we see that $\mathop {\lim }\limits_{n \to \infty } \dfrac{{{a_n}}}{{{b_n}}} = 1$ , which is a positive, defined value.
According to the limit comparison test this tells us that $\sum {a_n}$ and $\sum {b_n}$ are either both convergent or both divergent.
Since ${b_n} = \left( {\dfrac{1}{n}} \right)$ , we see that $\sum {b_n}$ is divergent (it's the harmonic series), so we can conclude that $\sum {a_n} = \sum\limits_{n = 1}^\infty {\sin \left( {\dfrac{1}{n}} \right)}$ is also divergent.

Note : In order to apply this limit comparison test test we need both series to start at the same place.
Do not misuse this test.
the requirement that ${a_n},{b_n} \geqslant 0$ and ${a_n} \leqslant {b_n}$ really only needs to be true eventually. In other words, if a couple of the first terms are negative or ${a_n}\;is not \leqslant {b_n}$ for a couple of the first few terms we’re okay. As long as we eventually reach a point where ${a_n},{b_n} \geqslant 0$ , and ${a_n} \leqslant {b_n}$ for all sufficiently large n the test will work.