I am trying to show the following sum is bounded:
$$ \sum_{k=1}^\infty \frac{\left(e^{ik} + \sin(\sqrt 2 k) + \cos(\sqrt 3 k)\right)^3}{k}$$
and to show that the magnitude
$$\left | \sum_{k=1}^\infty \frac{\left(e^{ik} + \sin(\sqrt 2 k) + \cos(\sqrt 3 k)\right)^3}{k} \right| < 6$$
So, as for what I've done so far. Consider just the numerator each of the sums $\left | \sum_k e^{ik} \right |$, $\sum_k \sin({\sqrt 2 k})$, $\sum_k \cos({\sqrt 3 k})$. Notice there exists $\delta \in \mathbb R$ for which each sum is bounded within some $[-\delta, \delta]$. (That is not exactly trivial but there are some resources online explaining those).
You can see this easily by the monoticity of $1/k$, $\lim_{k\to\infty}1/k=0$, and the boundedness of the sums above, that the sum written without a cube certainly converges (Dirichlet's test).
To solve the cube, we just need to apply the same logic with a larger bound. Products of numbers magnitude $[-\delta, \delta]$ are contained within $[-\delta^2, \delta^2]$, and similarly, the product of any three of these must be contained in $[-\delta^3, \delta^3]$. Now, the largest binomial coefficient is 6, We have overall that the sum indeed is within $[-48\delta^3, 48\delta^3]$. Therefore the sum certainly converges and hence the magnitude in question exists.
Now for calculating the upper bound. Should we try to upper bound each component of the binomial sum? Or do we need to use some trick of mixing $e^{ik}$ with irrationally paired since and cosine waves. It seems like a tricky problem, and I'm stuck getting it within 6. Certainly I could say with $\delta=3$ (I'm sure we could find tighter) that the sum is at most 432. Perhaps I should just come up with bullet points of techniques to reduce the sum and hope they can just continue to combine and combine till finally pushing the sum below 6.
Any help would be appreciated – Thank you..
Best Answer
The idea here is to turn the series into a bunch of smaller series whose values are known.
The standard product-to-sum identities: $$ \cos(a)\cos(b) = \frac{1}{2}\left(\cos(a+b)+\cos(a-b)\right)\\ \sin(a)\sin(b) = \frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)\\ \sin(a)\cos(b) = \frac{1}{2}\left(\sin(a+b)+\sin(a-b)\right) $$
Next, we have two closed-form identities: $$ \sum_{k=1}^{\infty} \frac{\sin(M k)}{k} = \frac{\pi-(M\bmod 2\pi)}{2}\\ \sum_{k=1}^{\infty} \frac{\cos(M k)}{k} = -\frac{1}{2}\log(2-2\cos(M)), $$provided $\cos(M)\ne 1$, a condition we don't have to worry about as $\{1,\sqrt{2},\sqrt{3}\}$ are linearly independent and $\pi$ is transcendental over $\mathbb{Q}$. With this in mind, we can take a deep breath and expand the numerator and get a sum of harmonic sines and cosines; the exact result, which contains a few terms, is here. At any rate, each individual series converges, and the exact value of the original sum can then be found. It is quite gory (too large to include in this post; too slow for WolframAlpha to compile on the free plan) but there is a closed-form that approximates to: $$ z = \sum_{k=1}^{\infty} \frac{\left(e^{ik}+\sin(\sqrt{2}k)+\cos(\sqrt{3} k)\right)^3}{k}\approx -0.799762+5.93444i $$Indeed, we have $|z|\approx 5.98809$, so the size of the series is very slightly less than $6$, as you hoped.