$x+\frac{x^2}{2}+\frac{x^3}{3}+..+\frac{x^n}{n}=\ln n$ has an unique positive root $x_n$ Prove that the equation: $x+\frac{x^2}{2}+\frac{x^3}{3}+..+\frac{x^n}{n}=\ln n$ has an unique positive root $x_n$, and find the value of 
$$\lim_{n\to \infty}x_n^{\sqrt{\ln n}}$$
Here, $x_n$ is the unique root of the equation for each n, which I figured out. But I still couldn't finish it.. Please help me
Sorry, I've made a mistake in the question.. So sorry you guys :( 
 A: For $0<x\le a<1$, $$x+\frac{x^2}2+\cdots+\frac{x^n}n<\ln\frac1{1-x}\le\ln\frac1{1-a}.$$
The first inequality follows because it is an equality in the limit as $n\to\infty$.
With $a=1-\frac1n$ we conclude from this that $x_n>1-\frac1n$. We also have $x_n<1$, so $x_n\to1$ as $n\to\infty$.
Now we turn our attention to the desired limit. Taking the logarithm, we want the limit of $\sqrt{\ln n}\ln x_n$ as $n\to\infty$. Since $x_n\to1$, we get $\ln x_n\sim x_n-1$, or more precisely
$$ \lim_{n\to\infty} \frac{\ln x_n}{x_n-1}=1,$$
so we might as well compute the limit of $\sqrt{\ln n}(x_n-1)$ instead.
From the first paragraph we get
$$ -\frac{\sqrt{\ln n}}{n}<\sqrt{\ln n}(x_n-1)<0. $$
Taking the limit as $n\to\infty$, we get $0$ on the left hand side, and so by the squeeze law $\sqrt{\ln n}(x_n-1)\to0$, and therefore $\sqrt{\ln n}\ln x_n\to0$ as well. Now take the exponential, and get $x_n^{\sqrt{\ln n}}\to1$.
(Note: This answer went through some revisions, as the question got changed along the way.)
A: Just a hint
Let for $n>1$
$$f_n (x)=\sum_{k=1}^n\frac {x^k}{k}-\ln (n) $$
we have
$$f (0)=-\ln (n)<0$$
$$\frac {1}{k+1}<\ln (k+1)-\ln (k)<\frac {1}{k} $$
by sum
$$\ln (n+1)<f_n (1)+\ln (n) $$
thus $$f_n (1)>0$$
$f $ is continuous at $[0,1] $ hence by IVT, it has a root $x_n $ at $(0,1) $.
$$f'_n (x)=1+x+...x^n/n>0$$
$x_n $ is unique.
A: Let $P_n(x)$ be the given polynomial. Just from staring down areas of rectangles, we have
$$\tag 1 P_n(1)-1 \le \ln n \le P_n(1).$$
Now Taylor's theorem shows that for $x<1,$
$$\tag 2 P_n(x) = P_n(1) + P'(1)(x-1)+P''(c_x)(x-1)^2/2$$ $$ \le P_n(1) + P'(1)(x-1)+P''(1)(x-1)^2/2,$$
where we used the fact that $P''$ is increasing. Simple calculations show $P'(1) = n,$ $ P''(1) = (n-1)n/2.$ Thus $(2)$ says
$$P_n(x) \le P_n(1) + n(x-1)+(n-1)n(x-1)^2/4.$$
I just played around with this last inequality to see that if $x=1-2/n,$ then $P_n(x) \le P_n(1) -1.$ Recalling $(1),$ we see that $1-2/n\le x_n \le 1.$ Thus
$$(1-2/n)^{\sqrt {\ln n}} \le x_n^{\sqrt {\ln n}} \le 1.$$
It's easy to see the term on the left $\to 1,$ hence the desired limit is $1$ by the squeeze theorem.
