Is this function an injection of $\mathbb{N} \to \mathbb{R}$?

Question

This comes from an earlier question of mine. It seems unapproachable.

Is the arithmetic function $$f(n) := \sum_{k=1}^n \frac 1{\sin(k)}$$ An injection?

I think the answer is yes.

Answer 1

Yes, this is an injection - this may be solved by observing that $$\sin(k)=\frac{z^k-z^{-k}}{2i}$$ where $z=e^i$. The importance of this transformation is that $z$ is transcendental, hence two rational functions of $z$ with rational (or, generally, algebraic) coefficients are equal if and only if they are equal as rational functions. If we toss out the coefficient of $2i$, since its a constant multiplied by every term, we then find that the question asks whether the sums $$\sum_{i=1}^n\frac{1}{z^k-z^{-k}}$$ are all distinct at $z=e^i$. This is then equivalent to asking whether these summands are equal everywhere as functions of $z$ since "transcendental" means that any polynomial or rational equation with rational coefficients that is true for $z$ is true for every number - but clearly no pair of these functions are equal everywhere because the $n^{th}$ term has a pole at every ${2n}^{th}$ root of unity, whereas no prior function has this property.