`sum_(n=1)^oo sin(1/n)` Use the Limit Comparison Test to determine the convergence or divergence of the series.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

Limit comparison test is applicable when `suma_n` and `sumb_n` are series with positive terms. If `lim_(n->oo)a_n/b_n=L` where L is a finite number and `L>0` , then either both series converge or both diverge.

Given series is `sum_(n=1)^oosin(1/n)`

Let the comparison series be `sum_(n=1)^oo(1/n)`

The comparison series `sum_(n=1)^oo1/n` is a p-series of the form `sum_(n=1)^oo1/n^p` with p=1.

p-series test states that `sum_(n=1)^oo1/n^p` is convergent if `p>1` and divergent if `0<p<=1`

So ,the comparison series is a divergent series.

Now let's use the limit comparison test with:`a_n=sin(1/n)`

and `b_n=1/n`



Let's apply L'Hopital's rule to evaluate the limit.

Test L'Hopital's condition: `0/0`







Since the comparison series `sum_(n=1)^oo1/n` diverges,so the series `sum_(n=1)^oosin(1/n)` as well ,diverges by the limit comparison test. 

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial