Proving ∑k=1n|cos⁡(k)|≥n4 for all n&gt;0I can do it using the fact that |cos⁡(x)|≥cos⁡(x)2∑k=1n|cos⁡(k)|≥∑k=1ncos⁡(k)2=n2+12Re(e2i1−e2∈1−e2i)=n2+12Re(ei(n+1)sin⁡(n)sin⁡1)=n2+cos⁡(n+1)sin⁡(n)2sin⁡(1)≥n2−12sin⁡(1)

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You do not need great accuracy for&amp;nbsp;sin⁡1. In fact, to shown2−12sin⁡1≥n4it suffices to haven≥2sin⁡1From&amp;nbsp;π&amp;lt;4, we have&amp;nbsp;sin⁡1&amp;gt;sin⁡π4=122, and hance are doen for all&amp;nbsp;n≥22, so (as&amp;nbsp;2&amp;lt;94=32) certainly for all&amp;nbsp;n≥3. A high precision calculator would not have helped you better than that.For n=2, the claim is&amp;nbsp;|cos⁡1|+|cos⁡2|≥12&amp;nbsp;and for n=1, it is&amp;nbsp;|cos⁡1|≥14. This time we use&amp;nbsp;1&amp;lt;π3, hence&amp;nbsp;cos⁡1&amp;gt;cos⁡π3=12

Proving \(\displaystyle{\sum_{{{k}={1}}}^{{n}}}{\left|{\cos{{\left({k}\right)}}}\right|}\ge{\frac{{{n}}}{{{4}}}}\) for all n>0 I can do

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