# 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?

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 ${a}_{n} = \sin \left(\frac{1}{n} ^ 2\right)$

Let's think about the end behaviour of ${a}_{n}$.

For large $n$, the argument $\frac{1}{n} ^ 2$ becomes small.

We can use the small-angle approximation:

As x→0, sin x → x.

So, for large $n$, ${a}_{n}$ acts like $\frac{1}{n} ^ 2$.

Let ${b}_{n} = \frac{1}{n} ^ 2$.

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 $x = \frac{1}{n} ^ 2$

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

But ${b}_{n} = \frac{1}{n} ^ 2$ is convergent, so

${a}_{n} = \sin \left(\frac{1}{n} ^ 2\right)$ is also convergent.