# Is $\{\sum_{k=1}^n\sin(k)\}_{n=1}^\infty$ bounded? [duplicate]

Is $$\{\sum_{k=1}^n\sin(k)\}_{n=1}^\infty$$ bounded?

More precisely, do there exist $$M_1,M_2 \in \mathbb{R}$$ such that $$M_1<\displaystyle\sum_{k=1}^n \sin(k) for every $$\ n\in\mathbb{N}\ ?$$

Would it be possible to use a variant of the integral test for convergence, but instead of making it about convergence, make it about boundedness?

Maybe there is some more generalised theorem that can be used to also answer the naturally modified question of the above one:

Given $$\ a,b\in\mathbb{R},\$$ does there exist $$\ M_1,M_2 \in \mathbb{R},\$$ such that $$M_1<\displaystyle\sum_1^n \sin(a+bn) for every $$\ n\in\mathbb{N}\ ?$$

I know that the terms of the sequence $$\ a_n=\sin(n)\$$ or $$\ b_n=\sin(an+b)\$$ that we are summing, are dense in $$\ [-1,1]\$$, but I don't see how this is helpful in answering the question(s).

Does this has something to do with Ito Calculus or Stochastic Calculus (which I know nothing about, by the way)?

• Yes, the partial sums involving $\sin(n)$ are bounded. Jun 20 at 16:32
• Really? Cool. Well, I'm interested in a proof or the type of maths required for a proof. Jun 20 at 16:34
• @AdamRubinson Use $\sin (n) = \Im (e^{in})$ and the geometric series. See also math.stackexchange.com/questions/1119043/…
– Gary
Jun 20 at 16:35

Instead of summing just the sines, imagine that you sum vectors with both sines and cosines: $$\sum_{n=0}^k \begin{bmatrix}\cos(n) \\ \sin(n)\end{bmatrix}$$ Then what you're looking for is the $$y$$-coordinate of the result.

The reason to include the cosines in the $$x$$-coordinates is that now each partial sum is the sum of a series of unit-length vectors, each turned by $$1$$ radian relative to the previous one.

This means that the partial sums for all $$k$$ will lie on some circle in the plane, with a radius just of the right size that a chord spanning an angle of $$1$$ radian has length $$1$$. And the $$y$$ coordinates of points on that circle are obviously bounded.

This reasoning also applies to summing $$\sin(a+bn)$$ as long as $$b$$ is real and not a multiple of $$2\pi$$.

(Figuring out the exact center of the circle is an interesting exercise, but is not necessary for simply concluding that there is a bound).

• Very nice approach. Jun 20 at 18:09
• I don't think this argument makes the result as transparent as it claims to. For each individual circle you are talking about, it is clear that the $y$-coordinates on that particular circle are bounded because we are talking about a unit circle somewhere in the plane. What is less clear (at least to me) is that (some subsequence of) the $y$-coordinates of the centers of the individual unit circles aren't moving to $\pm\infty$ as we add unit vectors with these incremental turns of $1$ radian, and this is what needs to be ruled out to conclude the partial sums of the original series are bounded. Jun 20 at 18:28
• @AlexOrtiz: As long as we keep $a$ and $b$ constant, there's only one circle (though it is not a unit circle), and each partial sum is a point on that particular circle, offset from the previous partial sum by one radian of arc. Jun 20 at 18:30
• So it is the existence of that circle that isn't clear to me. Ostensibly you use that the difference in the angles is $1$. Maybe this is obvious, but I don't see it straightaway. Jun 20 at 18:34
• @AlexOrtiz: Let $(0,0)$ and $(1,0)$ be the first two partial sums (namely of zero and one term, respectively). Their difference is $(\cos 0,\sin 0)$. Now let the line between them be the base of an isosceles triangle whose apex angle is $1$ radian. If you rotate that triangle by 1 radian about the apex, one corner moves to the other's position, and the vector corresponding by the base is turned by 1 radian. So you get a new point at $(\cos 0+\cos 1, \sin 0+\sin 1)$, and if you rotate it by another radian you get to $(\cos 0+\cos 1+\cos 2,\sin 0+\sin 1+\sin 2)$. Jun 20 at 18:39

Alternatively to rephrase the other answer in complex form, consider $$e^{ik} = \cos(k) + i \sin(k)$$. Then $$\sum_{k=1}^n \sin(k)$$ is the imaginary part of $$z_n=\sum_{k=1}^n e^{ik}=\frac{e^i(1-e^{in})}{1-e^i}$$ as we have a finite geometric sum. Now the norm of this complex number $$z_n$$ is bounded by $$\left|\frac{e^i(1-e^{in})}{1-e^i}\right|\le \frac{2}{|1-e^i|}$$ And since the imaginary part of a complex number is no greater than the norm of the complex number, we have $$\left|\sum_{k=1}^n \sin(k)\right|\le|z_n|\le\frac{2}{|1-e^i|}.$$ (Note $$e^i \neq 1$$, so the fraction is finite.)