Answer to Question #264344 in Calculus for jimmi

Question

Convergence test for $\displaystyle\sum_{n=1}^\infty \frac{sin(n)}{n}$.

Answer 1

Let us show that the series $\sum\limits_{n=1}^{\infty}\frac{\sin(n)}n$ is divergent.

Note that $|\sin x|⩾\frac{1}2$ for $x∈I_k=[\frac{π}6+kπ,π−\frac{π}6+kπ].$

Let us find the length of every $I_k:$

$|I_k|=π−\frac{2π}6=\frac{2π}3>2.$

We conclude that every $I_k$ contains at least one natural number $n_k.$

Then

$\sum\limits_{n=1}^N\frac{|\sin(n)|}n ⩾\frac{1}2\sum\limits_{n_k\le N}\frac{1}{n_k}\to\infty$ when $N→∞$

since the harmonic series $\sum\limits_{n=1}^{∞}\frac{1}n$ diverges.