TITLE: Convergence of $\sum\limits_{n = 1}^{\infty} \sin^2(\pi/n)$
QUESTION [2 upvotes]: 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?

REPLY [0 votes]: 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.
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