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

Question

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?

Answer 1

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

Answer 2

$x+\sin\, x$ 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 $x \geq 0$.

Answer 3

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.

Answer 4

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 $$ \sin x=\sum_{k=0}^\infty (-1)^m\frac{x^{2m+1}}{(2m+1)!},~~\cos x=\sum_{k=0}^\infty (-1)^m\frac{x^{2m}}{(2m)!}. $$

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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 $s_{2n}$.

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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 $x<0$.