Prove that series with $(-1)^k\cos^3(k)$ has sequence of partial sums bounded.

Question

I found that $(-1)^k\cos^3(k)$ equals $$(1-sin^2(k))cos((\pi+1)k)$$ or $$cos((\pi+3)k) + sin(k)sin((\pi+2)k) + sin(k)cos(k)sin((\pi+1)k)$$.

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 $cos(\pi k)$ and the trigonometric identities.

Answer 1

$$ \begin{align} &\left|\,\sum_{k=0}^{n-1}(-1)^k\cos^3(k)\,\right|\\ &=\left|\,\sum_{k=0}^{n-1}\frac{(-1)^k}8\left(e^{ik}+e^{-ik}\right)^3\,\right|\\ &=\left|\,\sum_{k=0}^{n-1}\frac{(-1)^k}8\left(e^{3ik}+3e^{ik}+3e^{-ik}+e^{-3ik}\right)\,\right|\\ &\le\frac18\left|\,\frac{1-\left(-e^{3i}\right)^n}{1+e^{3i}}\,\right|+\frac38\left|\,\frac{1-\left(-e^{i}\right)^n}{1+e^{i}}\,\right|+\frac38\left|\,\frac{1-\left(-e^{-i}\right)^n}{1+e^{-i}}\,\right|+\frac18\left|\,\frac{1-\left(-e^{-3i}\right)^n}{1+e^{-3i}}\,\right|\\ &\le\frac1{4\cos\left(\frac32\right)}+\frac3{4\cos\left(\frac12\right)} \end{align} $$

Answer 2

The partial sum sequence will look like $a_n= \sum_{k=1}^{n}(-1)^kcos^3(k)$ Now write, $4cos^3(k)=cos(3k)+3cos(k)$. Then try to use the triangle inequality.

Answer 3

Hint: By the triple angle formula, $(\cos \theta)^3$ is a linear combination of $\cos (3\theta)$ and $\cos \theta$.

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For each $\alpha,\beta$ the partial sums of $\sum_{k=1}^\infty \cos(k\alpha+\beta)$ are bounded. (It is the difference of geometric series $\sum \exp(i(k\alpha+\beta)) - \sum \exp(-i(k\alpha+\beta))$.)

Apply this to $\sum \cos(2k)$, $\sum \cos(2k+1)$ for the series $\sum (-1)^k \cos k$ and to $\sum \cos(6k)$, $\sum \cos(6k+3)$ for the series $\sum (-1)^k \cos (3k)$