Thue–Morse sequence ($t_n=s_2(n)\mathrm{mod}2$)

Here $s_2(n)$ is the sum of the binary digits of the nonnegative integer $n$. The sign version is $f(n)=(-1{)}^{s_2(n)}$. Concatenating binary digits gives $f(a{2}^j+r)=f(a)f(r)$ for $0\le r<2^j$. It supplies cancellation controlled by binary blocks rather than multiplicative factorization.

Each binary digit independently contributes either $1$ or $-e(2^l\theta )$, proving the product. Two adjacent factors have modulus $8|\cos (\pi t)|(1-{\cos }^2(\pi t))\le 16/(3\sqrt{3})<4$. Consequently the product is $O(2^{\sigma j})$, with $\sigma =2-\frac{3}{4}{\log }_{2}3<1$. This two-factor estimate suffices for a power saving, though its exponent is not optimal.

An integer interval is a disjoint union of aligned binary blocks with at most two of each length. On $[a{2}^j,(a+1)2^j)$ the Thue–Morse sequence sign factors as $f(a)f(r)$, so the Thue–Morse Fourier product bounds its exponential sum by $O(2^{\sigma j})$. Sum the geometric series over block lengths to obtain the uniform bound, for $\sigma =2-\frac{3}{4}{\log }_{2}3$.

The additive character orthogonality filter $1_{m\mid r}=m^{-1}{\sum }_{a=0}^{m-1}e(ar/m)$ restricts the uniform interval cancellation for binary digit parity to multiples of $m$, without increasing its bound. Each inner Type I sum is therefore $O(N^{\sigma })$. Summing the at most $N^{1/100}$ outer terms proves the displayed estimate, which is smaller than every fixed negative power of $\log N$ for large $N$.