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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 $\alpha \gt \frac12$

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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.

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