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

i'm very puzzled, what stops you from generating the sequence? or using a graph calculator? isn't it just using a for loop? — Siong Thye Goh, Apr 09 '19 at 13:17
@SiongThyeGoh Thanks for giving direction, but never did it so. I think you mean that $n$ is integer. Am sure that $n$ can be taken as real too, with a defined degree of precision in terms of no. of decimal digits. The python code here (https://www.geeksforgeeks.org/program-calculate-value-sinx-cosx/) should work after modifn. in value of input $x$, & print(sin(x)) rather than print(round(sinx)). — jiten, Apr 09 '19 at 14:01
... don't write your own sine function... call some library. also for integer $n$, $\sin(n \pi/2)$ takes only $3$ values. — Siong Thye Goh, Apr 09 '19 at 14:05
@SiongThyeGoh For such simple exercise, would not prefer library, & the same concern is shared here too : https://stackoverflow.com/q/45169675/3693431, although the selected answer there uses library only. — jiten, Apr 09 '19 at 14:29
programming sine values and this maths questions are two separate questions. — Siong Thye Goh, Apr 09 '19 at 15:15
@SiongThyeGoh Request elaboration, as unable to understand your last comment. — jiten, Apr 09 '19 at 15:46
@SiongThyeGoh Sorry, was not on system. Can we join chat now? — jiten, Apr 10 '19 at 01:11

score 2 · Answer 1 · answered Apr 09 '19 at 10:16

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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.

answered Apr 09 '19 at 10:16

Peter Foreman

19,947

kindly detail the usage of squeeze theorem here. – jiten Apr 09 '19 at 10:17
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We have that $\lim_{n\to\infty} \pm \frac1n=0$ so the term $a_n$ is bounded by $0\le a_n\le 0$ in the limit. I.e. $\lim_{n\to\infty} a_n =0$ – Peter Foreman Apr 09 '19 at 10:19

score 1 · Answer 2 · answered Apr 09 '19 at 10:15

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$ 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.

answered Apr 09 '19 at 10:15

Kavi Rama Murthy

311,013

Siong Thye Goh · Accepted Answer · 2019-04-09T17:49:21.847

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 $0$.

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

edited Apr 09 '19 at 17:49

answered Apr 09 '19 at 14:03

Siong Thye Goh

149,520
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Kindly see my post at : https://math.stackexchange.com/q/3195527/424260 – jiten Apr 21 '19 at 06:10
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I'm traveling again. The other link should have answered the question. You should be dealing with abstract set rather than intervals – Siong Thye Goh Apr 21 '19 at 06:17
Please see edited post, & my last comment. Attempt Q.3: quadratic $ax^2+bx+c$, with $a=3, b=-10, c=3$, roots $x=\frac13, 3$. Form a bounded set taken by $x$ in $R$, in which range $\lt 0, {x \in R: \frac13 < x < 3}$. So, no max, but min, supermum. Coming to linear factors: $(3x - 1)(x - 3)$ in bounded domain, with supremums
$8, 0$ resp. at $x=3$. While at $x = \frac13$, have supremum values in range as $0, \frac{-2}3$. The product of supermums at each end = $0$ too as needed. --- Better if chat was there. – jiten Apr 22 '19 at 00:54
Please see my post at : https://math.stackexchange.com/q/3196871/424260 – jiten Apr 22 '19 at 09:29
If you want to add something that will help me further, please see my last two comments' posts. – jiten Apr 22 '19 at 16:47
Kindly help with my post at : https://math.stackexchange.com/q/3201343/424260, It is about book question on supremum, bounds. – jiten Apr 25 '19 at 04:20
i might attempt after work if no one answered it (which is unlikely). – Siong Thye Goh Apr 25 '19 at 05:12
There is one comment only. Although that comment has corrected my error in one row, but still worried about four more rows. – jiten Apr 25 '19 at 05:24
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I think there are multiple mistakes of which I can only answer after work. – Siong Thye Goh Apr 25 '19 at 05:26

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

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