Chebyshevs inequality convergence of higher moments Let $X_1,..,X_N $ be random variables with expectation value denoted by $ \mu_i$. Assume that all moments exist. Using chebyshevs inequality for higher moments we get.
$$ \mathbb{P} (\frac{1}{N} \sum_{i=1}^N (X_i - \mu_i) \geq \epsilon) \leq \frac{\mathbb{E} [( \frac{1}{N} \sum_{i=1}^N (X_i - \mu_i))^m]}{\epsilon^m} $$
for all $m \geq 2$. My question is now. If I can show that 
$$ \frac{\mathbb{E} [( \frac{1}{N} \sum_{i=1}^N (X_i - \mu_i))^2]}{\epsilon^2} \underset{ N \rightarrow \infty} \longrightarrow 0 $$
does this imply that $$ \frac{\mathbb{E} [( \frac{1}{N} \sum_{i=1}^N (X_i - \mu_i))^m]}{\epsilon^m} \underset{N \rightarrow \infty} \longrightarrow 0 $$
for all $m \geq 2$. I think this must hold but I dont get it. In my opinion this could only hold if $(X_i - \mu_i)^m$ decreases i.e. $(X_i - \mu_i)^m \leq (X_i - \mu_i)^{m+1}$ for all $m \geq 2$. For me it would be sufficient if  $(X_i - \mu_i)^m \leq (X_i - \mu_i)^{m+2}$
 A: This is not necesarily true. Now, I'll show a counter-example to this proposition (btw, I'm sorry, but the counter-example is somewhat convoluted)
Let $\epsilon < 1$, and define, for all $n$, the random variable $Z_n$, s.t.
$$
\Bbb{P}[Z_n=x]=
\begin{cases}
\frac{1}{n^3}&x=\pm n\\
1-\frac{2}{n^3}&x=0\\
0 & \text{otherwise}
\end{cases}
$$
for which we have $\Bbb{E}[Z_n]=0$
Now, notice that for all $n$,
$$
\Bbb{E}[|Z_n|^2]=n^2\cdot \frac{1}{n^3}+(-n^2)\cdot\frac{1}{n^3}=\frac{2}{n}
$$
thus $\Bbb{E}[|Z_n|^2]\to 0$ as $n\to\infty$
On the other hand, 
$$
\Bbb{E}[|Z_n|^4]=n^4\cdot \frac{1}{n^3}+(-n^4)\cdot\frac{1}{n^3}=2n
$$
thus $\Bbb{E}[|Z_n|^4]\to \infty$ as $n\to\infty$
To see why this has any relevance to our case, let $X_1=\epsilon Z_1,\: X_2=\epsilon(2Z_2-Z_1),\: X_3=\epsilon(3Z_3-2Z_2),\:...$, and, in general, $X_{n+1}=\epsilon((n+1)Z_{n+1}-nZ_n)$.
We can show$$
\text{for all $n$},\: \Bbb{E}[X_n]=\mu_n=0
$$
and
$$
Z_n = \frac{\frac{1}{n}\sum_{i=1}^n X_i}{\epsilon}=
\frac{\frac{1}{n}\sum_{i=1}^n (X_i-\mu_i)}{\epsilon}
$$
In other words, we have found a sequence of random variables $X_1,...,X_n,..$ s.t. have all their moments, and for which
$$
\Bbb{E}\left[\frac{\left(\frac{1}{n}\sum_{i=1}^n(X_i-\mu_i)\right)^2}{\epsilon^2}\right]=\Bbb{E}[|Z_n|^2]\to 0
$$
but
$$
\Bbb{E}\left[\frac{\left(\frac{1}{n}\sum_{i=1}^n(X_i-\mu_i)\right)^4}{\epsilon^4}\right]=\Bbb{E}[|Z_n|^4]\to \infty$$
