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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?

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  • $\begingroup$ Use the limit comparison test with the harmonic series. $\endgroup$ – Adam Hughes Feb 25 '17 at 16:39
  • $\begingroup$ If you know how to do the estimates on the sine function, that works just fine, but taking limits is required for either version, and doing direct comparison means you have to take the extra step of showing the limit being one implies the inequality given. I'm not poo-pooing a good direct comparison: I love that test the best, but I would say it's "more basic" rather than "easier." $\endgroup$ – Adam Hughes Feb 25 '17 at 18:46
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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.

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Use equivalents:

$\sin\dfrac1n\sim_\infty\dfrac1n$, which diverges, hence the series $\displaystyle\sum_{n\ge1}\biggl\lvert\sin\dfrac1n\biggr\rvert$ diverges.

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$\frac{\sin(1/n)}{1/n} \to 1$ 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 $n>N$.

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Hint: if $x$ is small, then $\sin(x)$ is very close to $x$.

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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.

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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.

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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.

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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 $\lim_{n\to \infty} H_n = \infty$

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