How do you use the limit comparison test to determine if #Sigma sin(1/n)# from #[1,oo)# is convergent or divergent?

Let #a_n=sin(1/n)# and #b_n=1/n#.

Then #lim_(nrarroo)a_n/b_n=lim_(nrarroo)sin(1/n)/(1/n)#. There are multiple ways to approach this limit. The first is to replace the variables: as #nrarroo,1/nrarr0#, so this can be rewritten as #lim_(ararr0)sin(a)/a#, which is a fundamental limit: #lim_(ararr0)sin(a)/a=1#. So:

#lim_(nrarroo)sin(1/n)/(1/n)=lim_(ararr0)sin(a)/a=1#

Another way of going about the limit is to apply L'Hopital's rule, since it's in the indeterminate form #0/0#. So:

#lim_(nrarroo)sin(1/n)/(1/n)=lim_(nrarroo)(-1/n^2cos(1/n))/(-1/n^2)=lim_(nrarroo)cos(1/n)=cos(0)=1#

Either way, we see that #lim_(nrarroo)a_n/b_n=1#, which is a positive, defined value.

According to the limit comparison test this tells us that #suma_n# and #sumb_n# are either both convergent or both divergent.

Since #b_n=1/n#, we see that #sumb_n# is divergent (it's the harmonic series), so we can conclude that #suma_n=sum_(n=1)^oosin(1/n)# is also divergent.

#sinx = x - x^3/(3!)+x^5/(5!)+cdots# is an alternating series and
#x-x^3/(3!) < sinx < x# for #-pi le x le pi# so

#1/n-1/(6n^3) lt sin(1/n) lt 1/n# but

#sum_(k=1)^oo1/n-1/(6n^3)# diverges because

#sum_(k=1)^oo-1/(6n^3)# converges (#approx -0.2#) and

#sum_(k=1)^oo1/n# diverges