Growth rate of Taylor convergents near pole For any fixed $z_0\in\mathbb{C}\setminus \{0\}$ and $\beta\in\mathbb{R}^{+}$, prove that $$\left.T_n\left(\log^{\beta}z;z_0\right)\right|_{z=0}\sim\log^{\beta} n$$
Note: I observed that this holds when $\beta$ is natural, as well as an analogue for some other functions with poles at zero (in place of $\log$) like $z^{\alpha}$ for $\alpha<0$ (although here, the exponent sign flips).
$T_n(f(z);z_0)$ denotes the Taylor approximant of $f(z)$ around $z_0$ of order $n$.
 A: This is not true in general. When $\beta$ is fractional, the zeros of $\log z$ create additional singularities of the function, reducing the radius of convergence of its power series and thus changing the asymptotics of Taylor polynomials. For example, take $z_0=e$ and $\beta=1/2$. Writing $z=e-ew$ and using the principal branch of $\log$ (you did not specify the branch), we find that 
$$\log z = \log (e-ew)=1+\log(1-w)=1-\sum_{k=1}^\infty \frac{w^k}{k}\tag1$$ 
With $\beta=1/2$ the binomial series is
$$(1-\zeta)^{1/2}=1-\sum_{n=1}^\infty \frac{(2n-3)!!}{(2n)!!}\zeta^n\tag2$$
When we plug $\zeta=\sum_{k=1}^\infty \frac{w^k}{k}$ into (2), all coefficients of powers of $w$ will be negative on every occasion: there is no cancellation between them. 
Let's estimate the size of the coefficient of $w^{3m}$. It comes in part  from the term in (2) with $n=2m$, namely 
$$
-\frac{(4m-3)!!}{(4m)!!} (w+w^2/2+\dots)^{2m} 
\tag3$$
Indeed, the binomial expansion of (3) contains the term
$$
-\frac{(4m-3)!!}{(4m)!!} \binom{2m}{m} \frac{1}{2^{m}} w^{3m}   
\tag4$$
Estimate the central binomial coefficient from below: 
$$
\binom{2m}{m}  \ge \frac{2^{2m-1}}{\sqrt{m}}
\tag5$$ 
and observe that 
$$
\frac{(4m-3)!!}{(4m)!!} > \frac{1}{(4m+1)^2}\frac{(4m+1)!!}{(4m)!!} > \frac{1}{(4m+1)^2} 
\tag6$$
Putting (5) and (6) into (4), we see that the coefficient of $w^{3m}$ tends to infinity (exponentially) 
as $m\to\infty$. Since all coefficients are  negative, the value of Taylor polynomial at $w=1$ (which corresponds to $z=0$)
also grows exponentially. 
One could  avoid the estimates with binomial coefficients by observing that 
the function $f(w)=(1+\log(1-w))^{1/2}$ is not holomorphic in $|w|<R$ when $R>1-e^{-1}$ (because $w=1-e^{-1}$ 
is problematic). Therefore, the radius of convergence of its Taylor series at $w=0$ is at most $1-e^{-1}$, which implies that coefficients of $w^j$ grow exponentially at least along  some subsequence. 
