How can we control $\beta$ to make the mass of the n-dimensional unit ball concentrate in a layer? As a special phenomenon of high dimension, we know that for arbitary $\epsilon>0$,  as $n \rightarrow ∞$. we have: 
Where $m$ denotes the Jordan measure and $B_n$ denotes the $n$-dimensional unit ball. 
We say that the mass of the ball $B_n$ concentrates in an $O(n^α)$ boundary layer if 
Now I am asked to find a infimum for $\alpha$ to make the mass concentrates in an $O(n^α)$ boundary layer.
That is, find the $\beta$ defined as:
$\beta:=inf\lbrace α < 0 :$ Mass of $B_n$ concentrates in an $O(n^α)$ boundary layer.$\rbrace$. 
And once we find the $\beta$, may I ask what can we say about the concentration of mass if (1)$α < β$, (2)$ \alpha=\beta$, (3) $\alpha>\beta$?
I have not exposed to some convoluted high-dimensional geometry context and I find myself lost in this question so far. I searched but I cannot find any material which can lead me to a start point. May I please ask for an answer (together with some explaination) if possible? Or any help or reference would be appreciate. Thanks in advance!
 A: The question boils down to evaluating the limit and see for which values of $\alpha$ does the limit evaluates to 1.
First we have to know how the formula for $m(B_n(1-n^\alpha))$ relates to $m(B_n)$. Since the n-balls of radius $1$ and $(1-n^\alpha)$ and are similar, their volumes are proportional to the $n^{th}$ power of their radii.
 In particular, $$\frac{m(B_n(1-n^\alpha))}{m(B_n)} = \left(\frac{(1-n^\alpha)}{1}\right)^n $$ (Think about how areas are proportional to the square of the ratio of radii; volumes in 3D are proportional to the cube of the ratio, etc.)
Then
$$\frac{m(B_n \backslash B_n(1-n^\alpha))}{m(B_n)} = \frac{m(B_n)-m(B_n(1-n^\alpha)}{m(B_n)} =1-(1-n^\alpha)^n$$
We want to find out for which $\alpha$ this limit tends to 1 as $n \to \infty$, in other words, the questions is for which $\alpha$, $$\lim \limits_{n\to\infty} (1-n^\alpha)^n = 0?$$
Now it's time for some algebra.
$$\lim \limits_{n\to\infty} (1-n^\alpha)^n = \lim \limits_{n\to\infty} \left(1-\frac{1}{n^{-\alpha}}\right)^n$$
$$=\lim \limits_{n\to\infty} \left(1-\frac{1}{n^{-\alpha}}\right)^{n^{-\alpha}\cdot n^{\alpha+1}} $$
$$=\lim \limits_{n\to\infty} e^{-n^{\alpha+1}}$$
The rest is left for you to figure out! For what $\alpha$ does the above expression evaluates to 0? 
And once you figure out the range of $\alpha$, the infimum $\beta$ is obvious. :)
As for the conditions (1)$\alpha < \beta$, (2)$\alpha = \beta$, and (3) $\alpha > \beta$ go, think about what the limit evaluates to for all of the above cases. What implications does that have for the original ratio that we were interested in, i.e. 
$\frac{m(B_n \backslash B_n(1-n^\alpha))}{m(B_n)}$? Is that still 1? Or other values? What does that mean geometrically?
Hope that helps! Good luck! 
