Does $\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right) $ absolutely converge? Does $$\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right)
$$ converge conditionally or absolutely?
I know that this series converges conditionally using the Leibniz's convergence test, but what method should be used to decide whether it converges absolutely?
 A: No, it does not converge absolutely. Note $\sin x \sim x$ for small $x$ and hence $\sin \frac 1n \sim \frac 1n$ for large $n$. This implies $\sin \frac{1}{n} \geq \frac{1}{2n} \geq 0$ for large $n$. But we know that $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n} =+\infty$ and so by comparison $\sum \sin \frac{1}{n} = + \infty$.
However, the series converges conditionally. This is an immediate consequence of the alternating series test. $\left|\sin \frac{1}{n}\right| = \sin \frac{1}{n} \to 0$ as $n \to \infty$ and $\sin \frac{1}{n}$ is positive and monotonically decreasing. 
A: Use equivalents:
$\sin\dfrac1n\sim_\infty\dfrac1n$, which diverges, hence the series $\displaystyle\sum_{n\ge1}\biggl\lvert\sin\dfrac1n\biggr\rvert$ diverges.
A: $\frac{\sin(1/n)}{1/n} \to 1$ for $n \to \infty$. Hence there is $N$ such that
$$\frac{\sin(1/n)}{1/n} \ge 1/2$$
for $n>N$. Therefore
$\sin(1/n) \ge \frac{1}{2n}$  for $n>N$.
A: Let $x_n=(-1)^n \sin {\frac 1n}$ and $y_n=\frac 1n$.
Then $\lim_{n \to \infty} |\frac {x_n}{y_n}|= 1 \neq 0.$
By Limit Comparison Test, since $\sum y_n$ is not absolutely convergent hence $\sum x_n$ also not absolutely convergent.
A: Hint: if $x$ is small, then $\sin(x)$ is very close to $x$.
A: It does not converge absolutely, because
$$\left\vert (-1)^n\sin\left(\frac 1n\right)\right\vert=\vert\sin(1/n)\vert$$
and 
$$\sin(x)\underset{x\to 0}\sim x$$
so
$$\sin(1/n)\underset{n\to\infty} \sim 1/n.$$
But 
$$\sum \frac 1n$$
diverges.
So it does not converge absolutely.
A: A series $\sum u_n$ converges absolutely if the series of its absolute values  $\sum |u_n|$ converges. Let us study the series $\sum_{n\geqslant 1} |(-1)^n \sin(1/n)|$ .
For $n \geqslant 1$, the ratio $1/n$ belongs to $]0,\pi/2[$ . Therefore, one deduces from the concavity of the sine function over $]0,\pi/2[$ that
\begin{equation}
\frac{2}{n\pi}<\sin\left(\frac{1}{n}\right)<\frac{1}{n} \, .
\end{equation}
Therefore, a partial sum satisfies
\begin{equation}
\frac{2}{\pi} \sum_{n=1}^N \frac{1}{n}<\sum_{n=1}^N \left|(-1)^n \sin\left(\frac{1}{n}\right)\right|<\sum_{n=1}^N \frac{1}{n} \, .
\end{equation}
Since the sequence of the partial sums of the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ increases to infinity, so does the sequence of the partial sums of the present series. The studied series does not converge absolutely.
Note: only the left-hand side of the inequalities is required.
A: As an addition to this problem, I want to say some particular case which I teach students.
It includes some way of other thinking and especially for positive series and so absolute convergence. 
$0<\lim_{n\to \infty} n\sin(\frac{1}{2n})=\frac{1}{2} <\lim_{n\to \infty} \sum_{N=n}^{2n} \sin(\frac{1}{n})$
Since that, the limit of the series is $\infty$.
Other case in which we can prove divergence of series.
$H_{n} = \sum_{k=1}^{n} \frac{1}{k}$
$H_{2n} - H_{n} > \frac{1}{2}$
$\min\text{d}(H_{2n},H_{n}) \neq 0 >\frac{1}{2}$
then $\lim_{n\to \infty} H_n = \infty$
