Evaluating the limit for the Minkowski content of $F_{\alpha}=\{0,1,\frac{1}{2^{\alpha}},\frac{1}{3^{\alpha}},\frac{1}{4^{\alpha}},\dots\}$. The Minkowski content is defined as
$\displaystyle M_{\beta}(A)=\lim_{\delta \rightarrow 0} \frac{\mu(A_{\delta})}{(2\delta)^{1-\beta}}$
where $0 < \beta < 1$, $A \subset \mathbb{R}$, and $A_{\delta}$ is the $\delta$ - neighbourhood around $A$.
Evaluate $M_{1/(1+\alpha)}(F_{\alpha})$ where
$F_{\alpha}=\{0,1,\frac{1}{2^{\alpha}},\frac{1}{3^{\alpha}},\frac{1}{4^{\alpha}},\dots\}$.
I am struggling to see how this can be answered.
 A: If $(\delta_k)$ is a sequence of positive numbers tending to zero, then
$$
M_{1/(1+\alpha)}(F_\alpha)=\lim_{k\to\infty}(2\delta_k)^{-\alpha/(1+\alpha)}\mu(F_{\alpha,\delta_k}).
$$
Choose $2\delta_k=k^{-\alpha}-(k+1)^{-\alpha}$.
Then
$$
F_{\alpha,\delta_k}
=
(-\delta_k,(k+1)^{-\alpha}+\delta_k)\cup\bigcup_{i=1}^k(i^{-\alpha}-\delta_k,i^{-\alpha}+\delta_k).
$$
Note that these open intervals are disjoint; we have chosen $\delta_k$ so that $(k+1)^{-\alpha}+\delta_k=k^{-\alpha}-\delta_k$.
Now it is easy to evaluate the Lebesgue measure of $F_{\alpha,\delta_k}$:
$$
\mu(F_{\alpha,\delta_k})
=
[(k+1)^{-\alpha}+2\delta_k]+k\times2\delta_k
=
(k+1)^{-\alpha}+(k+1)(k^{-\alpha}-(k+1)^{-\alpha}).
$$
Therefore
$$
M_{1/(1+\alpha)}(F_\alpha)
=
\lim_{k\to\infty}[k^{-\alpha}-(k+1)^{-\alpha}]^{-\alpha/(1+\alpha)}[(k+1)^{-\alpha}+(k+1)(k^{-\alpha}-(k+1)^{-\alpha})].
$$
The problem has thus reduced to evaluating a limit of an explicit sequence.
Note that
\begin{eqnarray}
k^{-\alpha}-(k+1)^{-\alpha}
&=&
k^{-\alpha}(1-(1+1/k)^{-\alpha})
\\&=&
k^{-\alpha}(\alpha k^{-1}+O(k^{-2}))
\end{eqnarray}
and
\begin{eqnarray}
(k+1)^{-\alpha}
&=&
k^{-\alpha}(1+1/k)^{-\alpha}
\\&=&
k^{-\alpha}(1-\alpha k^{-1}+O(k^{-2})).
\end{eqnarray}
These give
\begin{eqnarray}
(k+1)^{-\alpha}+(k+1)(k^{-\alpha}-(k+1)^{-\alpha})
&=&
k^{-\alpha}(1-\alpha k^{-1}+O(k^{-2}))
\\&&
+(k+1)k^{-\alpha}(\alpha k^{-1}+O(k^{-2}))
\\&=&
k^{-\alpha}(1+\alpha+O(k^{-1})).
\end{eqnarray}
Also
\begin{eqnarray}
[k^{-\alpha}-(k+1)^{-\alpha}]^{-\alpha/(1+\alpha)}
&=&
[k^{-\alpha}(\alpha k^{-1}+O(k^{-2}))]^{-\alpha/(1+\alpha)}
\\&=&
\alpha^{-\alpha/(1+\alpha)}k^\alpha(1+O(k^{-1})).
\end{eqnarray}
Combining these estimates we get
\begin{eqnarray}
M_{1/(1+\alpha)}(F_\alpha)
&=&
\lim_{k\to\infty}[\alpha^{-\alpha/(1+\alpha)}k^\alpha(1+O(k^{-1}))][k^{-\alpha}(1+\alpha+O(k^{-1}))]
\\&=&
\lim_{k\to\infty}\alpha^{-\alpha/(1+\alpha)}(1+\alpha)(1+O(k^{-1}))
\\&=&
\alpha^{-\alpha/(1+\alpha)}(1+\alpha)
.
\end{eqnarray}
It is of course possible that I have made a mistake in the calculations somewhere.
If you find a mistake, let me know.
