Does $\sum_{n=1}^{\infty} \sin\left(\frac{\pi}{n}\right)$ converge? I'm not really sure how to start here, the basic tests don't work.
 A: $\sin(\pi/n) = \pi/n + O(1/n^2)$ for $n$ large. Now use the fact that the harmonic series diverges.
A: Remember the usual limit from basic calculus
$$
\lim_{x \to 0} \frac{\sin x}{x} = 1
$$
Now, can you think of a way to use this basic limit in combination with the limit comparison test to conclude something about the convergence or divergence of your series? Off course you'll have to take $a_n = \sin{\left( \dfrac{\pi}{n} \right)}$ as one of your sequences ;-)
A: Use the limit comparison test, with the series $b_n =1/n$. Since:
$$\lim_{x\to0} \frac{\sin(x)}{x} =1$$
And since $b_n$ diverges, so does $a_n = \sin(\pi/n)$.
A: Hint: Approximate the sine function by a straight line.  What if it were a line?  Then use the limit comparison test.
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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Since $\sin\pars{x} \geq 2x/\pi$ when $0 \leq x \leq \pi/2$:
\begin{align}
\sum_{n = 1}^{N}\sin\pars{\pi \over n}
&=
\sum_{n = 2}^{N}\sin\pars{\pi \over n}
\geq
\sum_{n = 2}^{N}{2 \over \pi}\,{\pi \over n}
=
2\sum_{n = 2}^{N}{1 \over n}
=
2\sum_{n = 1}^{N}{1 \over n} -1
\end{align}
The right hand side diverges when $N \to \infty$ !!!.
