The series $\sum_{n=1}^{\infty} \frac {\cos n}{2n^{\alpha}}$ converges for $\alpha \in (0,1)$.

Question

I'm trying to solve this problem: Let $\alpha>0$ and $a_{n}=\frac{\cos n}{2n^{\alpha}}$
for all $n\in\mathbb{N}$. Prove that the series $\sum_{n=1}^{\infty}a_{n}$
coneverges.

For $\alpha>1$, we have that $\mid\cos x\mid\leq1$ for all $x\in\mathbb{R}$
and therefore for all $n\in\mathbb{N}$:
$$
\left\lvert \frac{\cos n}{2n^{\alpha}}\right\rvert=\frac{\lvert\cos n\rvert}{2n^{\alpha}}\leq\frac{1}{2}\cdot\frac{1}{n^{\alpha}}
$$

Using the fact that $\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}$ is
convergent for $\alpha>1$, we have for the comparison test that $\sum_{n=1}^{\infty}a_{n}$
is absolutely convergent for $\alpha>1$. And with this $\sum_{n=1}^{\infty}a_{n}$
is convergent. But I don't know how can I prove that $\sum_{n=1}^{\infty}a_{n}$
converges when $0<\alpha<1$. Could you help me or give me some idea
to prove this?

Answer 1

Hint

Dirichlet’s test may be your friend.