Measure of a decreasing union of circles If we define $$A=\bigcup_{k=1}^{\infty}S\left(\frac{1}{k^\alpha}\right)$$ for $2>\alpha>1$ where $$S\left(\frac1{n^\alpha}\right)=\left\{(x,y):x^2+y^2=\frac{1}{k^{2\alpha}}\right\}$$
Then how would I show the following inequality 
$$0<M_{1}(A)<\infty$$
I understand that when we choose the neighbourhoods around each circle then at some point they will intersect and so I have arrived at the following 
$$\frac{\mu_2(V(\delta,A))}\delta=\sum_{k=1}^{n-1}4\pi\frac1{k^\alpha}+\frac{\pi(\frac1{n^\alpha}+\delta)^2}\delta$$
I cant seem to get a strict inequality from this though. I have tried to use approximations of sequences $\sum_{i=1}^n\frac1{i}$ and $\sum_{i=1}^n\frac1{i^2}$ to get an upper and lower bounds on the expression but have not arrived at the answer I want. 
 A: For $E \subset \mathbb R^2,$ $E_r$ denotes the set$\{z\in \mathbb R^2: d(z,E)<r\}.$  Note that if $E\subset F,$ then $E_r\subset F_r$ for all $r>0.$ Let $\lambda$ denote Lebesgue measure on $\mathbb R^2.$ As I understand it, the one-dimensional Minkowski content of $E\subset \mathbb R^2$ is
$$M_1(E)=\lim_{r\to 0^+} \frac{\lambda (E_r)}{2r},$$
provided the limit exists.
Fix $N\in \mathbb N.$ Then
$$\tag 1 \lambda\left ( \bigcup_{k=1}^{N} S(k^{-\alpha})_r\right) \le  \lambda(A_r).$$
If $r$ is small enough, then the sets $S(k^{-\alpha})_r$ are disjoint annuli whose measures are $\lambda (S(k^{-\alpha})_r)=4\pi r k^{-\alpha}.$ For such $r,$ the left side of $(1)$ equals $\sum_{k=1}^{N} 4\pi r k^{-\alpha}.$ Thus
$$\frac{\sum_{k=1}^{N} 4\pi r k^{-\alpha}}{2r}  \le \frac{\lambda (A_r)}{2r}.$$
This shows
$$ \sum_{k=1}^{N}2\pi k^{-\alpha}\le \liminf_{r\to 0^+} \frac{\lambda (A_r)}{2r}.$$
Now $N$ was arbitrary, so we can let $N\to \infty$ to get
$$\tag 2\sum_{k=1}^{\infty}2\pi k^{-\alpha}\le \liminf_{r\to 0^+} \frac{\lambda (A_r)}{2r}.$$
For an estimate on the other side, note that for any $r>0,$ $ A\subset \{|z|\le r\} \cup (\cup_{k^{-\alpha}>r} S(k^{-\alpha})).$ Thus
$$A_r \subset \{|z|\le 2r\} \cup (\cup_{k^{-\alpha}>r} S(k^{-\alpha})_r),$$
which implies
$$\lambda (A_r) \le 4\pi r^2 + \sum_{k^{-\alpha}>r}4\pi r k^{-\alpha} \le 4\pi r^2+\sum_{k=1}^\infty4\pi r k^{-\alpha}.$$
Therefore
$$\frac{\lambda (A_r)}{2r} \le 2\pi r +\sum_{k=1}^\infty2\pi  k^{-\alpha}\implies\tag 3\limsup_{r\to 0^+} \frac{\lambda (A_r)}{2r} \le \sum_{k=1}^\infty2\pi  k^{-\alpha}.$$
Together, $(2)$ and $(3)$ give  $M_1(A)=\sum_{k=1}^\infty2\pi  k^{-\alpha}.$
A: As you point out, we have:
$$
\frac{\mu_2(V(\delta,A))}\delta=\sum_{k=1}^{n-1}4\pi\frac1{k^\alpha}+\frac{\pi(\frac1{n^\alpha}+\delta)^2}\delta
$$
where the $n$ varies with $\delta$. Geometrically, you can see than $n$ will be the smallest $n$ such that $\frac{1}{n^\alpha} - \frac{1}{(n+1)^{\alpha}} < 2\delta$. Now observe that $\frac{\alpha}{(n+1)^{\alpha + 1}} < \frac{1}{n^\alpha} - \frac{1}{(n+1)^{\alpha}} < 2\delta$, which after some algebra means $n > \left(\frac\alpha{2\delta}\right)^\frac1{\alpha+1} - 1$, which implies
$$
\frac1{n^\alpha} < \frac{\delta^\frac\alpha{\alpha+1}}{\left((\frac\alpha2)^\frac1{\alpha+1} - \delta^\frac1{\alpha + 1}\right)^\alpha}
$$ 
Thus we have the upper bound:
$$
\frac{\mu_2(V(\delta,A))}\delta<\sum_{k=1}^{n-1}4\pi\frac1{k^\alpha}+\frac{\pi\left(\frac{\delta^\frac\alpha{\alpha+1}}{\left((\frac\alpha2)^\frac1{\alpha+1} - \delta^\frac1{\alpha + 1}\right)^\alpha}+\delta\right)^2}\delta = \sum_{k=1}^{n-1}4\pi\frac1{k^\alpha} + \pi\left(\frac{ \delta^{\frac{\alpha - 1}{2(\alpha + 1)}}}{\left((\frac\alpha2)^\frac1{\alpha+1} - \delta^\frac1{\alpha + 1}\right)^\alpha} + \delta^\frac12\right)^2
$$
Letting $\delta$ go to $0$, this gives us:
$$
M_1(A) =\lim_{\delta \rightarrow 0^+} \frac{\mu_2(V(\delta,A))}{2\delta} \le \sum_{k=1}^\infty 2\pi\frac1{k^\alpha} = 2\pi\zeta(\alpha)
$$
For a lower bound, note that
$$
\frac{\mu_2(V(\delta,A))}{2\delta} > \sum_{k=1}^{n-1} 2\pi \frac{1}{k^\alpha}
$$
Letting $\delta$ go to $0$ (which makes $n$ go to infinity), we find that $M_1(A)\ge 2\pi\zeta(\alpha)$. Hence $M_1(A) = 2\pi\zeta(\alpha)$, which is finite and positive for $\alpha > 1$.
