Estimating $\sum_{n=1}^N \cos(t\log n)$

Question

I've been thinking about the sum $$\sum_{n=1}^N \cos(t\log n)$$ for large $N$. If $t=0$ the case is obvious, but when $t\neq 0$ can this sum be bounded? I presume it diverges weaker than just $N$; probably something like $N^\epsilon$ for some $0<\epsilon<1$. Equivalently the sum $$\left|\sum_{n=1}^N e^{it \log n} \right|$$ could be considered. I've been thinking about Koksma, but not sure how to apply it here.

Any ideas?

Answer 1

Your guess is wrong. If it was right, then $t\log n$ would be equidistributed, according to Weyl's criterion, but it isn't.

See Show that $\{a \log n \}$ is not equidistributed for any $a$

Edit: on second reading, the question I posted doesn't have an answer and the connection is really not that obvious. So let's just prove it directly without reference to anything but basic calculus:

You can replace the sum by the integral

$$ \int_{1}^{N} \cos(t \log x) dx =\frac{1}{2} N (\cos \log N + \sin \log N) $$ with error $$ |\epsilon_N|\leq \sum^N |\cos'(\xi_n)| t/n $$ The integral grows like $N$ the error like $\log N$. This disproves your $N^{\epsilon}$ guess but falls just short of showing actual linear growth.

Answer 2

Since the above sum is quite famous (though the interesting part of it happens when $t$ is large wr $N$, say $t \ge N^2$ when appropriately formulated bounds on it are equivalent to the Lindelof Hypothesis) let's show the easy result that if we fix $t \ne 0$, then for any $N$ there is a $M(N,t)$ s.t.

$S_{N,M}=\sum_{n=N}^{N+M-1} e^{it \log n}$ satisfies:

$|S_{N,M(N,t)}| \ge \frac{N}{2t+2}, N > t+1$,

which immediately implies that no uniform bound $|S_N|=\left|\sum_{n=1}^N e^{it \log n} \right| \le C(t,f)Nf(N), f(N) \to 0$ can exist for any function $0<f(N) \to 0, N \to \infty$ or if you want in shorthand that $S_N=o(N)$ cannot hold for fixed $t$ and $N \to \infty$

Fix $t>0$ as the sum for negative $t$ is obtained by conjugation. Integrating $f(x)=x^{it-1}$ on the interval $N, N+k$, $|e^{it\log(N+k)}-e^{it\log N}| \le \frac{k(t+1)}{N}$, so $||S_{N,M}|-M|\le \frac{(t+1)M(M-1)}{2N} \le \frac{N}{2(t+1)}$for $M(N,t)=\frac{N}{t+1}, N >t+1$, so $M>1$

But this clearly implies the stated result:

$|S_{N,M(N,t)}| \ge \frac{N}{2t+2}, N > t+1$