How to prove $a_n = \frac1 n \sin(\frac12 n\pi)$ is a convergent sequence?

Question

In the book by Bromwich, it is stated that in the above sequence, there are infinitely number of terms may equal limit; hence it is a convergent sequence.

I mean that there is no mention that the sequence is oscillatory.

Hence, want to know the graph of the series.
Better if could be given some sort of python\c\c++ code to generate the sequence.That would help for all such

Answer 1

Notice that whenever $n$ is even, $a_n=0$, hence if the sequence converges, it better converges to $0$.

Usually $n$ is reserved for interger, in particular in this context when we use it as a sequence index. Hence if you are really intested, just write a for loop to generate a few numbers.

But suppose even if you treat it as a real number and you ask the question, does

$$\lim_{x \to \infty}\frac{\sin (nx)}{x}$$

exists? and if so, what is the limit (it better be zero if it exists for the same reason).

We can again, use squeeze theorem since $-1\le \sin(\pi x/2)\le 1$.

For large $x$, $x$ is positive $$-\frac1x \le \frac{\sin(\pi x/2)}{x}\le \frac1x$$

Now, by squeeze theorem, the limit is $0$.

$$0=\lim_{x \to \infty }-\frac1x \le \lim_{x \to \infty}\frac{\sin(\pi x/2)}{x}\le \lim_{x \to \infty}\frac1x=0$$

Answer 2

$ 0 \leq |a_n| \leq \frac 1 n$ and hence $a_n \to 0$ by Sqeeze Theorem. Also $a_n$ equals its limit whenever $n$ is even.

Answer 3

As $$-1\le\sin{(x)}\le 1$$ We have that $$-\frac1n \le a_n \le \frac1n$$ Hence $$\lim_{n\to\infty} a_n=0$$ By squeeze theorem.