Hello!
What is the best way to determine if a sequence like converges or diverges?
And how do we find its limit (the sin(n^2) on the numerator confuses me). Thanks in advance...
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
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.
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
Remember if the corresponding series converges it must converge to 0.
In your example
Not only does sin(n^2) converge it must converge to 0
You can always show a sequence converges to 0 by showing the corresponding series converges.
However in practice I've found it not to be the best first choice.
One notable exception is the sequence an = e^n/ (n!)