Question

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?

Answer 1

`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.