Let us show that the series n=1∑∞nsin(n) is divergent.
Note that ∣sinx∣⩾21 for x∈Ik=[6π+kπ,π−6π+kπ].
Let us find the length of every Ik:
∣Ik∣=π−62π=32π>2.
We conclude that every Ik contains at least one natural number nk.
Then
n=1∑Nn∣sin(n)∣⩾21nk≤N∑nk1→∞ when N→∞
since the harmonic series n=1∑∞n1 diverges.
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