Is $\sum\limits_{n=1}^\infty \sin{\frac{(-1)^{n+1}}{n}}$ convergent?

Question

$$ \mbox{Is}\quad \sum_{n=1}^\infty \sin\left(\left[-1\right]^{n + 1} \over n\right) \quad\mbox{ convergent ?.} $$

$$ \mbox{I know that }\quad\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\quad \mbox{is convergent.} $$

$\frac{\sin x}x $ goes to 1 as x goes to $0$. So I feel that the series should be convergent, but I can't prove it rigorously.

Answer 1

Yes. Sine is an odd function, so your summation is just:

$$\sum_n (-1)^n\sin \left({1\over n}\right)$$

which converges by the alternating series test.

Answer 2

Hint:

$$\sum_{n=1}^{\infty}\sin\left(\frac{(-1)^{n+1}}{n}\right)=\sum_{n=1}^{\infty}(-1)^{n+1}\sin\frac{1}{n}$$

Answer 3

Hint

Use the Taylor's formula:

$$\sin x=_0 x+\mathcal O(x^3)$$ and you find that the given series is the sum of two convergent series: one by the Leibniz theorem and the second by comparison with a convergent Riemann series.

Answer 4

Since $$\sum_{n=1}^\infty\sin\frac{(-1)^{n+1}}{n}=\sum_{n=1}^\infty(-1)^{n+1}\sin\frac{1}{n},$$ this is a alternate series. From the monotonicity (decreasing) of $\sin\frac{1}{n}$ and the limit $$\lim_{n\to\infty}\sin\frac{1}{n}=0,$$ we see that the series is convergent.