What is the estimation for the positive root of the following equation
$$
ax^k = (x+1)^{k-1}
$$
where $a > 0$ (specifically $0 < a \leq 1$). 


Could you point out some reference related to the question?



Root estimation
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1). 


I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The [Lagrange inversion formula](http://en.wikipedia.org/wiki/Formal_power_series#The_Lagrange_inversion_formula) provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds  

$$x=a^{-1}+(k-1)+ \sum\_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\, ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < k ^ {-k}(k-1) ^ {k-1} $. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below. If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients. 


II. [edit] For large positive values of $a$, we can write the equation in the form 
$$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \, .$$
Again, the Lagrange inversion formula provides a power series expansion of the inverse of the RHS; computing the radius of convergence this time we find $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating that inverse at $a^{-\frac{1}{k}}$ we get
$$x=\sum\_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\, ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)


III. It remains the case $ a := k^{-k} (k-1)^{k-1} $, though it should be covered as a limit case by the above series.


