Does $\sum_{n=1}^\infty(-1)^n \sin \left( \frac{1}{n} \right) $ absolutely converge?

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

Use equivalents:

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

Answer 5

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

Answer 6

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.

Answer 7

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.