Transcendence measure of \(e^{1/n}\) (Q6611638)

Let \(\alpha\) be a transcendental number and let \(k\) and \(H\) be a positive integers. Denote \(\omega(\alpha,k,H)\) the infimum of the numbers \(r>0\) such that \(\mid\sum_{j=0}^k a_j\alpha^j\mid>\frac 1{H^r}\) for all \((a_0,\dots ,a_k)\in\mathbb Z^{k+1}\setminus \{ \overline 0\}\) and \(\max_{i=0,\dots ,k}\mid a_i\mid \leq H\). Then the authors prove that if \(k\geq n\geq 2\) then \(\omega(e^{\frac 1n},k,H)\leq k+\frac {k^2\log k}{\log\log H}d(k)\) for each \(h\) with \(\log H\geq (k+n)\log^2(k+n)e^{(k+n)\log^2(k+n)}\) and \(d(k)=1.596\) for \(k=2\), \(d(k)=1.069\) for \(k=3\), \(d(k)=1.051\) for \(k=4\) and \(d(k)=(1+\frac {0.69}{\log k-1})\) for \(k\geq 5\).\N\NFor the entire collection see [Zbl 1542.11001].