Convergence of $\sum ( \cos \sqrt[3]{n^3 + \sqrt n + 7} - \cos \sqrt[3]{n^3 - 2\sqrt n + 3})$ I have some problem with this example: $$\displaystyle \sum_{n=2}^{\infty}\Bigg(\cos\Big(\sqrt[3]{n^3+\sqrt{n}+7}\Big) -\cos\Big(\sqrt[3]{n^3-2\sqrt{n}+3}\Big)\Bigg)$$
the only idea that crossed my mind is to use that $\cos x-\cos y=-2\sin\big({\frac{x+y}{2}}\big)\sin\big({\frac{x-y}{2}}\big)$ but later I don't know what to do with sines how to compare them or what else I can do with them ? 
 A: Can you see why $x-y= O(n^{-3/2})$?
When $z$ is small, $\sin z \approx z$.
Then use the comparison test with a $p$-series.
A: Methodology


*

*Write $\cos x - \cos y = -2 \sin \frac{x+y}{2}\sin \frac{x-y}{2}$.

*Disregard the $-2$ and the $\sin \frac{x+y}{2}$, as $\lvert \sin \alpha \rvert \leq 1$

*Instead, focus on the $\sin \frac{x-y}{2}$ term. Intuitively, as $x \approx y$ when $n$ is large, we should expect this to get small, and therefore for this sine term to get small.

*Expand $x-y$ in power series to see that $x-y = \frac{1}{n^{3/2}} + \frac{4}{3n^2} + \dots$

*As $\sin x \approx x$ when $x$ is small, conclude that $\sin \frac{x-y}{2}$ behaves like $\frac{1}{2n^{3/2}}$

*Conclude that your series converges.

A: Hint. You may write, as $n \to \infty$,
$$ \sqrt[3]{n^3+\sqrt{n}+7}=\left(n^3+\sqrt{n}+7\right)^{1/3}=n\left(1+\frac{\sqrt{n}+7}{n^3}\right)^{1/3}=n+\mathcal{O}_1\left(\frac{1}{n^{3/2}}\right)$$
and
$$ \sqrt[3]{n^3-2\sqrt{n}+3}=\left(n^3-2\sqrt{n}+3\right)^{1/3}=n\left(1+\frac{-2\sqrt{n}+3}{n^3}\right)^{1/3}=n+\mathcal{O}_2\left(\frac{1}{n^{3/2}}\right)$$
thus, using  $\displaystyle \cos x-\cos y=-2\sin{\frac{x+y}{2}}\cdot \sin{\frac{x-y}{2}}$, we get
$$\left| \cos\sqrt[3]{n^3+\sqrt{n}+7} -\cos\sqrt[3]{n^3-2\sqrt{n}+3}\right|\leq2 \left|\sin \left(\mathcal{O}_3\left(\frac{1}{n^{3/2}}\right)\right)\right|=\mathcal{O}_4\left(\frac{1}{n^{3/2}}\right)$$ and conclude that the initial series converges as does $\displaystyle \sum \frac{1}{n^{3/2}}$.
