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The tail of $p$-series ($p > 1$) can be upper bounded by (from e.g. here): \begin{equation} \sum_{j=n}^\infty \frac{1}{j^p} = O(n^{1-p}). \end{equation}

I'm interested in finding the upper bound of \begin{equation} \sum_{j=n}^\infty \frac{\sin(cj)}{j^p},~~ \text{ or } ~~ \sum_{j=n}^\infty \frac{\cos(cj)}{j^p}, \end{equation} where $c \in \mathbb{R}$ and $p = 2$?

Since $\lvert \sin(cn) \rvert, \lvert \cos(cn) \rvert\leq 1$, it's trivial that the original bound $O(n^{-1})$ holds. So my question is whether one can obtain a tighter bound, e.g. $O(n^{-2})$ or so.

user185671631
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Yes, you can obtain the $O(n^{-2})$ bound for each sum for almost all values of $c$, the edge case being $c=2\pi n,\ n\in\mathbb{Z}\ $ in which case $\sin(cj)=0$ and $\cos(cj)=1$ giving $O(n^{-1})$. From now I'll assume that $c\not= 2\pi n$ for convenience.

Firstly, it's useful to know that the sum of $\sin(cn)$ on its own is bounded by a constant depending on $c$, remember we already handled the edge case. You can show this using $\sin(x)=(e^{ix}-e^{-ix})/2i$ and the formula for a geometric series, but a proof is also given here.

To take advantage of this fact, we use partial summation. If you define

$$S_c(n) := \sum_{k=1}^n{\sin(cn)} = O_c(1)$$

or equivalently using $\cos(cn)$, then for $N\ge n$ we have

$$\sum_{j=n}^N{\frac{\sin(cj)}{j^2}} = \sum_{j=n}^N{\frac{S_c(n)-S_c(n-1)}{j^2}}$$

$$= \sum_{j=n}^N{\frac{S_c(n)}{j^2}} - \sum_{j=n-1}^{N-1}{\frac{S_c(n)}{(j+1)^2}}\tag{re-indexing to make the $S_c(n)$'s the same}$$

$$= \frac{S_c(n-1)}{n^2} + \frac{S_c(N)}{(N+1)^2} + \sum_{j=n}^N{S_c(n)\left(\frac{1}{j^2} - \frac{1}{(j+1)^2}\right)}$$

$$= O_c\left(\frac{1}{n^2}\right) + O_c\left(\frac{1}{N^2}\right) + \sum_{j=n}^N{O_c\left(\frac{1}{j^3}\right)} = O_c(n^{-2}) + O_c(N^{-2})$$

Letting $N\to\infty$ gives you the result. Although I skipped the sum containing $\cos(cj)$, using $\cos(x) = (e^{ix}+e^{-ix})/2$ and the same method as above you should get the same result.

Remark: The same process can be done with pretty much any nice function, not necessarily $j^p$ however, Abel Summation is probably more suited for this as it uses integration which will give the bound directly.

Luca Armstrong
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  • Thank you for the reply. Does $S_c(n)$ (either defined with $\sin$ or $\cos$) depend on $c$ so that we can further upper bound the leading term? – user185671631 Feb 19 '24 at 02:23
  • Yes sorry, I didn't go into much detail. You can always bound $S_c(n)$ by a constant depending on $c$ as long as $c\not=2\pi n$, but no better than $O(1)$. This is different for the edge case. I've updated my answer so hopefully it makes more sense. – Luca Armstrong Feb 19 '24 at 12:37
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    If you want an explicit bound for $S_c(n)$, using the geometric series a mentioned leaves you with $e^{ic}-1$ in the denominator, so it would look something like $$|S_c(n)| \le \frac{A}{|e^{ic}-1|}$$ where $A$ is an arbitrary constant. And there are nicer ways to write this by bounding the denominator – Luca Armstrong Feb 19 '24 at 12:42