Answer in Calculus for jimmi #264344

Convergence test for “displaystylesum_{n=1}^infty frac{sin(n)}{n}”.

Let us show that the series “sumlimits_{n=1}^{infty}frac{sin(n)}n” is divergent.

Note that “|sin x|u2a7efrac{1}2” for “xu2208I_k=[frac{u03c0}6+ku03c0,u03c0u2212frac{u03c0}6+ku03c0].”

Let us find the length of every “I_k:”

“|I_k|=u03c0u2212frac{2u03c0}6=frac{2u03c0}3>2.”

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

Then

“sumlimits_{n=1}^Nfrac{|sin(n)|}nnu2a7efrac{1}2sumlimits_{n_kle N}frac{1}{n_k}toinfty” when “Nu2192u221e”

since the harmonic series “sumlimits_{n=1}^{u221e}frac{1}n” diverges.