# Studying the convergence of the series $\sum_{n=1}^\infty|\sin\left(e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n} \right)|^\alpha$

Studying the convergence of the series $$\sum_{n=1}^\infty|\sin\left(e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n} \right)|^\alpha$$

I tried studying the convergence of the series but I didn't get the same result as in the solution.

Can somebody tell me if I did a mistake?

This is what I did. For $$\alpha \gt 0$$ $$|\sin\left(e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n} \right)|^\alpha \le |e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n}|^\alpha$$

Since

$$\lim_{n\to\infty}{\frac{e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n}}{\frac2n}}=0$$

This implies that $$|e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n}|^\alpha \le \left(\frac2n\right)^\alpha$$

Since $$\sum_{n=1}^\infty\left(\frac2n\right)^\alpha$$ converges for $$\alpha \gt 1$$ so does the $$\sum_{n=1}^\infty|\sin\left(e^{\frac{1}{n}}-\frac{2}{n}-\cos\sqrt\frac{2}{n} \right)|^\alpha$$.

For $$\alpha \le 0$$ since $$\lim_{n\to\infty}a_n \neq 0$$ then the series diverges.

In the solution, it says that the series converges for $$\alpha \gt \frac12$$

## 1 Answer

Your answer is not complete - write the Taylor series for the expression inside the sin in terms of $$\frac{1}{n}$$ and note that the free term and the first power cancel so you remain with terms of order 2 or higher, hence about $$\frac{1}{n^2}$$ and continue the way you did to conclude.