I found that (−1)kcos3(k) equals (1−sin2(k))cos((π+1)k) or cos((π+3)k)+sin(k)sin((π+2)k)+sin(k)cos(k)sin((π+1)k).
I wanted to flatten it into such a sum of terms that each has bounded partial sums, so the final sequence of partial sums is bounded too.
But I got the sin(k) and sin(k)cos(k).
I should use that (−1)k equals cos(πk) and the trigonometric identities.
Answer
|n−1∑k=0(−1)kcos3(k)|=|n−1∑k=0(−1)k8(eik+e−ik)3|=|n−1∑k=0(−1)k8(e3ik+3eik+3e−ik+e−3ik)|≤18|1−(−e3i)n1+e3i|+38|1−(−ei)n1+ei|+38|1−(−e−i)n1+e−i|+18|1−(−e−3i)n1+e−3i|≤14cos(32)+34cos(12)
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