Let $f\_n(x)=1+x+x^{\sqrt{2}}+x^{\sqrt{3}}+x^{\sqrt{4}}+\cdots+x^{\sqrt{n}}$ be a sequence of functions on the interval $[0, 1]$. Is there a good closed form approximation for such a function ( whatever "good" means), analogous to the well-known expression for the sum of geometric progression.


Such an approximation is needed to get an asymptotics on the positive root of equation $f\_n(x)=\ln(n)$ for large $n$.



Approximation of pseudogeometric progression
For natural $n$ and $x\in(0,1)$, one has
$$L\_n(x)\le f\_n(x)\le U\_n(x), \tag{10}\label{10}$$
where
$$U\_n(x):=1+I\_n(x), \tag{20}\label{20}$$
$$I\_n(x):=\int\_0^n x^{\sqrt t}\,dt
=2\frac{1+x^{\sqrt{n}} \left(\sqrt{n} \ln x-1\right)}{\ln^2 x},$$
$$L\_n(x):=I\_{n+1}(x). \tag{30}\label{30}$$


Next, fixing any real $c>0$ and letting $n\to\infty$, for $x=1-c/\sqrt{\ln n}$ we will have
$$\sqrt n\ln x
\sim-c\sqrt{n/\ln n}\to-\infty, $$
$$x^{\sqrt{n}}=\exp(\sqrt n\ln x)=\exp(-(c+o(1))\sqrt{n/\ln n}),$$
and hence
$$I\_n(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n.$$
Similarly (or because $1-c/\sqrt{\ln(n+1)}=1-(c+o(1))/\sqrt{\ln n}\;$), we have
$$I\_{n+1}(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n.$$


So, by \eqref{20} and \eqref{30},
$$U\_n(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n\sim L\_n(1-c/\sqrt{\ln n}).$$
Hence, in view of \eqref{10},
$$f\_n(1-c/\sqrt{\ln n})\sim\frac2{c^2}\,\ln n.$$


So, for the only zero $x=x\_n$ of the strictly increasing function $(0,1)\ni u\mapsto f\_n(u)-\ln n$ we have
$$x\_n=1-\sqrt{\frac{2+o(1)}{\ln n}}$$
as $n\to\infty$.


