Vtyjkyuk

Consider the function

$f_k(c):=sum_n=0^infty c^n^k$ where $kge 1$ is an integer. This one obviously converges for $leftlvert c rightrvert <1.$

In the following we study the solution to the equation

$$sum_n=0^infty left(n^k -frac1alpharight) c^n^k=0.$$

This one always exists as long as $alpha in (0,1).$

Numerically, I discovered something that I would like to understand:

As $alpha rightarrow 0$ we have that $c= 1-fracgammaalphak$ for some constant $gamma.$

So first the solution $c$ seems to depend in a linear way on $alpha$ for $alpha$ small and second, the dependence on $k$ also seems to be just $1/k$.

I would like to understand these two observations.

We can indeed show that for all $k$,
$$1-c sim_k fracalphak qquad (alpha to 0^+)$$

The middle part of my answer makes it intuitive why the main term is linear in $alpha$. What it does not explain, is whether there is a deeper reason that it is linear in $1/k$.

Uniqueness of the solution. Let's start by justifying that there is a unique solution $c in (0,1)$ for small $alpha$.
We have $$alpha = fracf_k(c)c cdot f_k'(c)$$
so we see that $alpha$, as a function of $c$, has a pole at $0$ of order $1$. In particular, $alpha(c) to +infty$ as $c to 0^+$.

In the sequel, we always assume $c in (0,1)$, and often write simply $f_k$ for $f_k(c)$. For the derivative we have $$alpha'(c) = fracc(f_k')^2 - f_k(cf_k')'(cf_k')^2$$
By Cauchy-Schwarz, $(f_k')^2 < c f_k(cf_k')'$ so that $alpha$ is strictly decreasing.

We have $alpha to 0$ for $c to 1^-$. Indeed, this follows e.g. by A Tauberian theorem for a quotient of power series, the limit on the boundary
There is probably a more elementary argument, but we will use the same result again anyway.

Using $alpha'<0$, $alpha > 0$, $alpha to +infty$ for $c to 0^+$ and $alpha to 0$ for $c to 1^-$, we conclude that $alpha : (0,1) to (0, infty)$ is a decreasing bijection.

Asymptotics. First take $k=1$, so that $f_k(c) = (1-c)^-1$ and we simply have $$alpha = frac1c-1$$

Let $k geq 1$ be arbitrary now. We will determine the limit of $alpha'$ as $c to 1$, but first finish the argument. Suppose the limit exists and equals $p_k neq 0$. Then $alpha$ has a $C^1$-extension to the right of the point $c=1$, and we have have $$alpha(c) = p_k cdot (c-1) + o_k(1) qquad (alpha to 0, c to 1^-)$$
The statement from the beginning of the post follows, with $p_k=-k$.

Limit of the derivative $alpha'(c)$.
We have
$$alpha' = 1- fracf(c)c g(c)$$
with $$f(c) := cf_k(cf_k')' = sum_m,n geq 0c^m^k+n^kn^2k =: sum_i geq 0a_ic^i$$
and $$g(c) := (cf_k')^2 = sum_m,ngeq 0c^m^k+n^km^kn^k =: sum_i geq 0b_ic^i$$
By the linked question, we only have to estimate the partial sums of the coefficients $a_n$ and $b_n$. We have $$A(N) := sum _i leq Na_i = sum_n^k+m^k leq N n^2k$$ and similarly for $B$. We can estimate the sum from below and above by a Riemann integral over domains of the form $$D_k, N = (x,y) = 0 leq x,y ;,; x^k + y^k leq N $$
Here are some nice pictures of what they look like: https://mathsci2.appstate.edu/~cookwj/maple/Circles/
It's not hard to see that
$$beginalign* A(N) &= int_D_k,N x^2k dxdy + O_k(N^2+1/k) \
&= int_0^N^1/k x^2k (N-x^k)^1/kdx + O_k(N^2+1/k) \
&= N^2+2/k int_0^1 u^2k (1-u^k)^1/kdu + O_k(N^2+1/k) endalign*$$
and similarly
$$B(N) = N^2+2/k frac1k+1int_0^1 u^k (1-u^k)^1+1/kdu + O_k(N^2+1/k) $$

We conclude that $$beginalign*
lim_N to inftyfracA(N)B(N) &= (k+1)fracint_0^1 u^2k (1-u^k)^1/kduint_0^1 u^k (1-u^k)^1+1/kdu \
&= (k+1) fracBleft(frac2k+1k,1+ frac1kright)Bleft(frack+1k, 2+frac1kright) \
&= k+1
endalign*$$
where $B$ is the Beta function, which satisfies $int_0^1 u^a(1-u^k)^bdu = B(fraca+1k, b+1)$ for $a,b > 0$.

Finally, $$p_k = lim_c to 1^-alpha'(c) = 1- lim_N to inftyfracA(N)B(N) = -k$$