Question #262920

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



1
Expert's answer
2021-11-15T02:21:48-0500

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

Note that sinx12|\sin x|⩾\frac{1}2 for xIk=[π6+kπ,ππ6+kπ].x∈I_k=[\frac{π}6+kπ,π−\frac{π}6+kπ].

Let us find the length of every Ik:I_k:

Ik=π2π6=2π3>2.|I_k|=π−\frac{2π}6=\frac{2π}3>2.

We conclude that every IkI_k contains at least one natural number nk.n_k.

Then

n=1Nsin(n)n12nkN1nk\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 NN→∞

since the harmonic series n=11n\sum\limits_{n=1}^{∞}\frac{1}n diverges.


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