# Convergence of $\sum\limits_{n = 1}^{\infty} \sin^2(\pi/n)$

I am trying to determine the convergence of $$\sum\limits_{n = 1}^{\infty} \sin^2(\pi/n)$$

After some time I found out that $$sin^2(\pi x) \leq (\pi x)^2$$ holds true for all $$x$$ using a graphing calculator. Which means I can substitute $$x={1\over n}$$ and get $$sin^2(\pi/n) \leq (\pi/n)^2$$. And since it is clear that $$\sum\limits_{n = 1}^{\infty}(\pi/n)^2$$ converges, $$\sum\limits_{n = 1}^{\infty} \sin^2(\pi/n)$$ converges as well by the comparison test.

The problem is how can I possibly know that $$sin^2(\pi x) \leq (\pi x)^2$$ holds true when I am taking an exam and I don't have enough time to mess around with my graphing calculator?

## 4 Answers

$$x+\sin\, x$$ and $$x-\sin\, x$$ are both non-decreasing functions since their derivatives are non-negative. They both vanish when $$x=0$$. Hence $$x\pm \sin \,x \geq 0$$ for all $$x \geq 0$$. This gives $$|\sin\, x| \leq x$$ and $$\sin^{2}x \leq x^{2}$$ for all $$x \geq 0$$.

• Thanks. Just one thing. I didn't understand the part where $a \pm b \geq 0$ gives $|b| \leq a$ – linearAlg Aug 15 at 5:25
• Oh I think I intuitively understood that statement. Thanks – linearAlg Aug 15 at 5:26
• $a\pm b \geq 0$ gives $-b \leq a$ and $b \leq a$. Since $|b|$ is either $b$ or $-b$ it follows that $|b| \leq a$. Also, please take a look at my comment for the other answer. – Kavi Rama Murthy Aug 15 at 5:28

Remember that $$0<\sin x for $$0 (which is the case for the terms in this question). Thus, squaring both sides and replacing $$x\mapsto\pi x$$ gives $$\sin^2\pi x<(\pi x)^2$$.

You are working with series, so I assume at exam time you also will know about power series. So at some point you will get to know that $$\sin x=\sum_{k=0}^\infty (-1)^m\frac{x^{2m+1}}{(2m+1)!},~~\cos x=\sum_{k=0}^\infty (-1)^m\frac{x^{2m}}{(2m)!}.$$

Directly related to series convergence is the Leibniz test for series $$\sum_{k=0}^\infty(-1)^ka_k$$, $$a_k>a_{k+1}>0$$ converging to $$0$$. One result of that test is that the value of that series is bounded by its partial sums $$s_n=\sum_{k=0}^n(-1)^ka_k$$, from below by the odd index sums $$s_{2m+1}$$ and from above by the even index sums $$s_{2n}$$.

Now in combination you get that the sine power series satisfies the Leibniz test if $$x^2<2k(2k+1)$$ for all $$k\ge1$$, that is, $$|x|<\sqrt 6$$. In consequence $$x-\frac{x^3}6\le\sin x\le x ~~ \text{ for } ~~ x\ge 0,$$ with the reverse relations for $$x<0$$.

This is really simple using asymptotic equivalence of functions:

Near $$0$$, $$\sin x \sim x$$, so $$\;\sin^2\dfrac\pi n\sim_\infty \dfrac{\pi^2}{n^2}$$, which is a convergent power series.

Now two series with equivalent general terms (and constant sign) both converge or both diverge.