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`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 `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`


`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. 

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