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

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 0So ,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
a_n/b_n=sin(1/n)/(1/n)
lim_(n->oo)a_n/b_n=lim_(n->oo)sin(1/n)/(1/n)
Let's apply L'Hopital's rule to evaluate the limit.
Test L'Hopital's condition: 0/0
=lim_(n->oo)(d/(dn)(sin(1/n)))/(d/(dn)(1/n))
=lim_(n->oo)(cos(1/n)(-1/n^2))/(-1/n^2)
=lim_(n->oo)cos(1/n)
lim_(n->oo)1/n=0
lim_(u->0)cos(u)=1
=1>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.