Question

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

Answer 1

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.

Answer 2

`sum_(k=1)^oo sin(1/k)` diverges

Explanation:

`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