How do you use the limit comparison test to determine whether the following converge or diverge given #sin(1/(n^2))# from n = 1 to infinity?

1 Answer
Aug 11, 2015

#color(red)(sum_(n=1)^∞ sin(1/n^2)" is convergent")#.

Explanation:

#sum_(n=1)^∞ sin(1/n^2)#

The limit comparison test states that if #a_n# and #b_n# are series with positive terms and if #lim_(n→∞) (a_n)/(b_n)# is positive and finite, then either both series converge or both diverge.

Let #a_n = sin(1/n^2)#

Let's think about the end behaviour of #a_n#.

For large #n#, the argument #1/n^2# becomes small.

We can use the small-angle approximation:

As #x→0, sin x → x#.

So, for large #n#, #a_n# acts like #1/n^2#.

Let #b_n= 1/n^2#.

Then #lim_(n→∞)a_n/b_n = lim_(n→∞)sin(1/n^2)/(1/n^2)= lim_(n→∞)(1/n^2)/(1/n^2) = 0/0#

This indeterminate result is discouraging, but we can apply L'Hôpital's rule:

If #lim_(x→a)f(x)/g(x) =0/0 or ±∞/∞#, then #lim_(x→a)f(x)/g(x)= lim_(x→a) (f'(x))/(g'(x))#.

Let #x = 1/n^2#

So, #lim_(x→0)sinx/x = lim_(x→0)cosx/1 =1/1=1#

But #b_n= 1/n^2# is convergent, so

#a_n = sin(1/n^2)# is also convergent.