# How to show that $\sin^2(1/n)\leq 1/n^2, \forall n\in\mathbb{N}?$

I'm trying to prove that $\sum_{n=1}^\infty \sin^2(1/n)$ is convergent, by using the comparison test.

I hypothesize that the sequence defined by $b_n:=1/n^2$ is always larger than $\sin^2(1/n)$. But when I do induction on $n$, the inequality isn't clear.

• Do you have to use the comparison test? I ask because the limit comparison test is very applicable here.
– user307169
Aug 22, 2017 at 0:21
• By using the same $b_n$? But then I can't get $a_n/b_n$ to converge. Aug 22, 2017 at 0:30
• Sure you can. See the answer I posted. Looks like someone else in the meantime posted a more full version also.
– user307169
Aug 22, 2017 at 0:39
• My edit was to repair an accidental mis-spelling of "comparison". Aug 22, 2017 at 3:05

Your hunch is correct, for all $x\in\mathbb{R}_{\geqslant 0}$, $\sin(x)\leqslant x$ as it can be shown using the derivative of $x\mapsto \sin(x)-x$ or the Taylor formula with integral remainder.

When deriving the derivative of $\sin$, you likely found the bounds

$$\cos(x)<\frac{\sin(x)}x<1\quad\forall~0<|x|<\pi/2$$

From this, you can easily see that

$$\sin(x)<x\quad\forall~0<x\le1\\\implies\sin^2(1/n)<1/n^2\quad\forall~1\le n$$

If you don't happen to remember this off the top of your head, it may be more intuitive to see that if $f''(x)<0~\forall~x\in(a,b)$, then $f(x)$ is bounded on that interval by it's tangent line and secant lines:

$$f(a)+\frac{f(a)-f(b)}{a-b}(x-a)<f(x)<f(a)+f'(a)(x-a)\quad\forall~a<x<b$$

Particularly,

$$\sin(x)<\sin(0)+\cos(0)x=x\quad\forall~0<x<\pi$$

Hint Use the Limit Comparison Test instead:

$$\lim_n \frac{\sin^2(1/n)}{1/n^2} =\left( \lim_n \frac{\sin(1/n)}{1/n} \right)^2=\left( \lim_{x_n \to 0} \frac{\sin(x_n)}{x_n} \right)^2=1$$

From the geometric definition of $\sin x:$

For $0<x<\pi /2,$ take triangle $AOB$ with $OA=OB=1$ and $\angle AOB=x.$ Let D lie on $OA$ with $DB\perp OA.$ The arc of the circle centered at $O,$ of radius $1,$ from $A$ to $B,$ has length $x,$ which is greater than the straight-line distance $AB.$ So $$\sin x=DB<\sqrt {DB^2+DA^2}\;=AB<x.$$

Since $\sin (-x)=-\sin x,$ therefore $|\sin x|<|x|$ for $0<|x|<\pi /2.$

For $|x|\geq \pi /2$ we have $|x|>1\geq |\sin x|.$

And of course for $x=0$ we have $|\sin x|=0=|x|.$

If you are allowed to use the limit comparison test, I recommend it. Use your same $b_n$. Hint to get you started:

$$\lim_{n\to+\infty} \frac{a_n}{b_n} = \lim_{n\to+\infty} \frac{\sin^2(1/n^2)}{1/n^2}$$

Now make the substitution $m = 1/n$. What does $m$ approach as $n \to +\infty$? The result of this substitution should be a familiar limit from Calc 1.

We have: $$|\sin x|=|\int_0^x\cos t dt|\le\int_0^x|\cos t| dt\le x$$ take $x=1/n$

Hence, $$\sum_{n=1}^\infty \sin^2(\frac1n)\le \sum_{n=1}^\infty\frac{1}{n^2}<\infty$$