Show that $|\sin\frac{1}{n^2}|<\frac{1}{n^2}$, $n=0, 1, 2, \dots$

Question

As part of showing that $$ \sum_{n=1}^\infty \left|\sin\left(\frac{1}{n^2}\right)\right| $$ converges, I ended up with trying to show that $$ \left|\sin\left(\frac{1}{n^2}\right)\right|<\frac{1}{n^2}, \quad n=1, 2, 3,\dots $$ since I know that the sum of the right hand side converges. But I can't show this. I've tried searching but I haven't been able to find anything.

What I've tried is that firstly, the absolute values are not needed since $\sin x>0$ if $0<x<1$. I rearranged a little bit: $$ \sin\left(\frac{1}{n^2}\right)-\frac{1}{n^2}<0 $$ and the derivative is $$ \frac{2}{n^3}\left(1-\cos\left(\frac{1}{n^2}\right)\right)> 0 $$ so my idea of showing that it is decreasing and negative for the first $n$ wouldn't work.

How can I show this? Help is appreciated.

Edit: Maybe I should add that I'm not completely sure it is true, but I tried it numerically and it seems like it.

The inequality $\sin x<x$ for $x>0$ was shown in several posts on this site. For example: How to strictly prove $\sin x<x$ for $0<x<\frac{\pi}{2}$, How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? — Martin Sleziak, Nov 29 '13 at 17:44
@MartinSleziak Yes, so I've realized. I think my problem is that I didn't make that quite obvious connection. This probably only proves that you should not study math on Friday afternoons :-) — hejseb, Nov 29 '13 at 18:16

score 3 · Accepted Answer · 2013-11-29T16:50:30.987

3

Consider $f(x)=\sin x -x\Rightarrow f'(x)=\cos x-1 < 0 \text { for x > 0 } \Rightarrow f(x)\text{ is strictly decreasing function}$

But then, $f(0)=0$ which would imply $f(x)< 0 $ for $x > 0$ i.e., $\sin x < x$ for $x >0$

Thus, $\sin \big(\frac{1}{n^2}\big)< \frac{1}{n^2}$

Does this imply $|\sin\big(\frac{1}{n^2}\big)|< \frac{1}{n^2}$

edited Nov 29 '13 at 16:50

answered Nov 29 '13 at 16:19

1

Nice, I know series expansion but I didn't really think of it. Possible dumb question: how do you know the remaining terms are less than 0 (when summed)? That second way is what my attempt was all about, but I'm not sure what went wrong... Thanks for showing me! – hejseb Nov 29 '13 at 16:27
Excuse me for late reply Some how i did not get any notice saying that you have commented..:O Oh yes.. I should have been more careful on the part that rest of the terms add up to positive number.... are you aware that I can see two elements in the series by combining if the series is more that just convergent? – Nov 29 '13 at 16:47
May be i am making too many assumptions (As i am not very sure of the proof at this point).. I would try making it more clear till then i would edit my answer with only one way... – Nov 29 '13 at 16:50

score 2 · Answer 2 · answered Nov 29 '13 at 16:00

It's actually very straightforward to show that $\sin x\leq x$ for $0<x<1$ which is all you have to do...

You can do it by definition using a unit circle.

You can do it by noting that $\sin 0=0$ and $\cos x<1$ here (i.e. $\sin x$ grows slower than $x$ here).

score 2 · Answer 3 · answered Nov 29 '13 at 16:02

2

If $0 < x < \pi/2$, $\sin(x)$ is the $y$--coordinate of the point on the unit circle distance $x$ via the circle. That is the shortest distance of any path from the point $P$, $(\cos(x), \sin(x))$ to the $x$--axis. Since $x$ describes the length of a path from $P$ to the $x$--axis that is not a straight line, we have $\sin(x) < x$. Draw of picture of this to see it nicely.

answered Nov 29 '13 at 16:02

ncmathsadist

49,383

Perhaps being more direct about what the path between $;P;$ and the $;x-$axis is (for the value $;x;$ can clear things out in case of need: it is the length of the (trigonometrix unit, say)circle's arc. Anyway, +1 . – DonAntonio Nov 29 '13 at 16:04

score 1 · Answer 4 · answered Nov 29 '13 at 16:02

1

Hint: $|\sin(t)| \le |t|$ for real $t$, with equality only at $0$.

answered Nov 29 '13 at 16:02

Robert Israel

448,999

3

I don't understand: isn't this almost-almost what the OP is asking about? – DonAntonio Nov 29 '13 at 16:05

Show that $|\sin\frac{1}{n^2}|<\frac{1}{n^2}$, $n=0, 1, 2, \dots$

4 Answers4