Does the series $\sum_{n=1}^{\infty}\sin\left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$ converge?

Question

Let $\alpha$ be such that $0\leq \alpha \leq 1$. Since $\sin n$ has no limit as $n$ tends to $\infty$, I'm having trouble with finding if the series $$\sum_{n=1}^{\infty}\sin \left(2\pi\sqrt{n^2+\alpha^2\sin n+(-1)^n}\right)$$ is convergent? Thanks.

Answer 1

This series is convergent.

As $n$ tends to $+\infty$, we may write $$ \begin{align} u_n &:=\sin \left( 2\pi \sqrt{n^2+\alpha^2 \sin n+(-1)^n}\right)\\\\ &=\sin \left( 2\pi n \:\sqrt{1+\frac{\alpha^2\sin n}{n^2}+\frac{(-1)^n}{n^2}}\right)\\\\ &=\sin \left( 2\pi n \:\left(1+\frac{\alpha^2\sin n}{2n^2}+\frac{(-1)^n}{2n^2}+\mathcal{O}\left(\frac{1}{n^4}\right)\right)\right)\\\\ &=\sin \left( 2\pi n +\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\\\ &=\sin \left(\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right)\right)\\\\ &=\frac{\pi\alpha^2\sin n}{n}+\frac{\pi(-1)^n}{n}+\mathcal{O}\left(\frac{1}{n^3}\right) \end{align} $$ Now recall that $\displaystyle \sum_{n=1}^{+\infty}\frac{\sin n}{n}$ is convergent, moreover $$ \sum_{n=1}^{+\infty}\frac{\sin n}{n}=\Im\sum_{n=1}^{+\infty}\frac{e^{in}}{n}=\Im\left(-\log(1-e^i)\right)=\frac{\pi-1}{2}. $$ Then it is clear that your initial series $\displaystyle \sum_{n=1}^{+\infty} u_n $ is convergent, being the sum of convergent series.

Answer 2

For simplicity, let $a_n = \alpha^2 \sin n + (-1)^n$. Notice that since $\sin$ has period $2\pi$,

$$\sin \left( 2\pi \sqrt{n^2 + a_n} \right) = \sin \left( 2\pi \sqrt{n^2 + a_n} - 2\pi n \right) = \sin \frac {4\pi^2 (\sqrt{n^2 + a_n})^2 - 4\pi^2 n^2} {2\pi \sqrt{n^2 + a_n} + 2\pi n} = \\ \sin \frac {4\pi^2 a_n} {2\pi \sqrt{n^2 + a_n} + 2\pi n} ,$$

therefore your series becomes

$$\sum \limits _{n \ge 1} \sin \frac {4\pi^2 a_n} {2\pi \sqrt{n^2 + a_n} + 2\pi n} .$$

Let us study the fraction inside the sine. Since

$$| a_n | = | \alpha^2 \sin n + (-1)^n | \le \alpha^2 + 1$$

and $\alpha$ is a constant, we deduce that the numerator is bounded. The denominator, on the other hand, tends to $\infty$, so the whole fraction will tend to $0$. In particular, there exist $n_0$ such that for $n \ge n_0$ the fraction will stay in $[0, \frac \pi 2]$, so its sine will be positive.

This allows us to use the limit comparison test to "drop" the sine, i.e. we deduce that the series has the same behaviour as

$$\sum \limits _{n \ge 1} \frac {4\pi^2 a_n} {2\pi \sqrt{n^2 + a_n} + 2\pi n} $$

(we use that if $x_n \to 0$, then $\dfrac {\sin x_n} {x_n} \to 1$).

Dropping those annoying constants and keeping in mind that $a_n$ is bounded, we can use the same test again to further simplify the denominator and obtain

$$\sum \limits _{n \ge 1} \frac {a_n} n .$$

Expanding this, we get

$$\alpha^2 \sum \limits _{n \ge 1} \frac {\sin n} n + \sum \limits _{n \ge 1} \frac {(-1)^n} n .$$

The first series is convergent by Abel-Dirichlet's test (this has been shown several times on MSE), while the second one is clearly convergent by Leibniz's test.