1. ## sine sequence

Hello!

What is the best way to determine if a sequence like $a_n = \frac{sin(n^2)}{n}$ converges or diverges?

And how do we find its limit (the sin(n^2) on the numerator confuses me). Thanks in advance...

2. A good trick that sometimes works when sines and cosines are involved is the squeezing thm

-1/n < sin(n^2)/n < 1/n

since lim-1/n and lim 1/n converge to 0 so does sin(n^2)/n

Makes Sense since you can think of the continuous curve sin(x)/x

as a sine curve whose amplitude 1/x decays to 0

3. So, the sequence converges to 0.

How come you chose to squeeze $sin(n^2)/n$ between ±1/n ? Why didn't you squeeze it between say, $- n^2/n$ and $n^2/n$ ?

4. The main reason is we have to squeeze it between 2 things which converge to the same limit. For sines and cosines their ranges are

-1 to 1 which is what makes the trick work.

yes -n^2/n < sin(n^2)/ n < n^2

But one limit is neg inf and the other pos inf

Which means we can only conclude the limit is somewhere between - inf and pos inf which means we know nothing about the limit.

5. can we use Cauchy's method

At first we check whether |sin(n^2) / n| converges or diverges ,

int {from 1 to oo} sin(x^2)/x dx
Sub u = x^2 => du = 2x dx
the integral becomes 1/2 int { from 1 to oo } sin(u) / u du
We know that int { from 0 to oo } sin(u) / u du converges to pi/2
so this integral 1/2 int { from 1 to oo } sin(u) / u du converges

Since the summation of |sin(n^2) / n| converge ,
sin(n^2) / n also converges too

6. Remember if the corresponding series converges it must converge to 0.