Give example of random variable which fulfills $\frac{M_n}{n^{1-\epsilon}}\to\infty$, where $M_n=\max_{i\in\{1,\dots,n\}}X_i$ I have difficulty with approaching following problem, because I am not sure if I understand everything correctly, so I put my reasining below with hope that it is not wildly incorrect.
Assume that $\epsilon>0$. Find sequence of random variables $X_i$ with same distribution such that $\mathbb{E}|X_i|<+\infty$ and $$\frac{M_n}{n^{1-\epsilon}}\to\infty$$ according to probability measure, where $M_n=\max_{i\in\{1,\dots,n\}}X_i$
$Z_n\to \infty$ according to probability measure when $\forall_{M>0}\exists_{n>0}\forall_{N>n} P(Z_N>M)=1$. In our case $Z_n=M_n/n^{1-\epsilon}$.
I guess we can assume that our variables are independent and surely $M_n/n^{1-\epsilon}\geq X_n/n^{1-\epsilon}$. Therefore we can look for sequence which will fulfill above condition for $Z_n=X_n/n^{1-\epsilon}$.
So let's set $M=n^\epsilon$, and we want $P(X_n>n)=1$. Am I correct so far?
If I am, then it seems that sequence $X_i$ with uniform distribution between $[i,i+1]$ would work.
 A: I will assume that you need that $Z_n\xrightarrow{p} +\infty$ (converges in probability). This is equivalent to 
$$
\forall M>0 \quad \mathbb P(Z_n\leq M)\to 0 \text{ as } n\to\infty. 
$$
Fix arbitrary $M>0$ and find this probability
$$
\mathbb P(Z_n\leq M) = \mathbb P\left(M_n\leq Mn^{1-\epsilon}\right) = \mathbb P\left(X_1\leq Mn^{1-\epsilon},\ldots,X_n\leq Mn^{1-\epsilon}\right) = \left(F_{X_1}\left(Mn^{1-\epsilon}\right)\right)^n,
$$
where $F_{X_1}(x)=\mathbb P(X_1\leq x)$ is the CDF of $X_i$. Rewrite the r.h.s. in terms of right tail $$\overline F(x)=1-F_{X_1}(x).$$
We get
$$\tag{1}\label{1}
\mathbb P(Z_n\leq M) = \left(F_{X_1}(Mn^{1-\epsilon})\right)^n = \left(1-\overline F\left(Mn^{1-\epsilon}\right)\right)^n
$$
This sequence tends to zero, if 
$$n\overline F\left(Mn^{1-\epsilon}\right)\to \infty \text{ as } n\to\infty.$$ 
On the other hand, the existence of first moment implies that $x\overline F(x)\to 0$ as $x\to +\infty$, and then
$$n^{1-\epsilon}\overline F\left(Mn^{1-\epsilon}\right)\to 0 \text{ as } n\to\infty.$$ 
So, we can get Pareto distribution with right tail $\overline F(x)=1/x^{1+\delta}$ ($x\geq 1$), where $\delta$ is choosen in such a way that $(1-\epsilon)(1+\delta)<1$. for example, let $\delta=\epsilon$. 
Then for the CDF
$$
F_{X_1}(x)=\begin{cases}0, & x<1\cr 1-\dfrac{1}{x^{1+\epsilon}}, & x\geq 1\end{cases}
$$
we have $\mathbb E[X_1]=1+\frac1\epsilon<\infty$, and (\ref{1}) looks as
$$
\mathbb P(Z_n\leq M) = \left(1-\frac{1}{M^{1+\epsilon}n^{(1-\epsilon)(1+\epsilon)}}\right)^n = \left(1-\frac{1}{M^{1+\epsilon}n^{1-\epsilon^2}}\right)^n \to 0 \text{ as } n\to\infty.
$$
