CLT for arithmetic mean of centred exp.distributed RV $X_1, X_2,\ldots$ are independent, exponentially distributed r. variables with $m_k:=EX_k=\sqrt{2k}$, $v_k:=\operatorname{Var} X_k=2k$.
I want to analyse the weak convergence of $Y_n=\dfrac{\sum_{i=1}^n (X_i−m_i)}{n}$.

CLT: When the Lindeberg Condition
$b^{−2}_n\sum_{i=1}^n E(|X_i−m_i|^2⋅1_{∣Xi−mi∣>ϵ⋅b_n}) ⟶0$ , for $n→∞\text{ and }∀ϵ>0$ is fullfilled, I can state
$\frac{\sum_{i=1}^n (X_i−m_i)}{b_n}⇒N(0,1)$. It converges weakly to a normal distributed r.v.
Here it is $b^2_n=\sum_{i=1}^n \operatorname{Var} X_i=\sum_{i=1}^n 2i=n(n+1)$
So if the condition would be fullfilled it is
$$Y_n=\frac{\sum_{i=1}^n (X_i−m_i)}{n}=\sqrt{\frac{n+1}{n}}\frac{\sum_{i=1}^n (X_i−m_i)}{b_n},$$
where
$$\frac{\sum_{i=1}^{n}(X_i−m_i)}{b_n}⇒N(0,1)\text{ and }\sqrt{\frac{n+1}{n}}=\sqrt{1+\frac{1}{n}}→1,$$
hence
$$Y_n⇒N(0,1).$$
Now the problem is to prove the Lindeberg condition.
$$E(|X_i−m_i|^2⋅1_{∣Xi−mi∣>ϵ⋅b_n})=∫_{∣x−m_i∣>ϵ⋅b_n}(x−m_i)^2\sqrt{2i}e^{−\frac{x}{\sqrt{2i}}}\;dx.$$
But thats where it ends. Am I on the right way for solving this? Can I switch ∑ and ∫ in the condition?

Let $x_{k,n}=\mathrm E((X_k-m_k)^2:|X_k-m_k|\geqslant\varepsilon b_n)$. Since $X_k/m_k$ follows the distribution of a standard exponential random variable $X$, 
$$
x_{k,n}=m_k^2\mathrm E((X-1)^2:(X-1)^2\geqslant\varepsilon^2 b_n^2/m_k^2).
$$
In particular, $x_{k,n}\leqslant m_k^2x_{n,n}$ for every $k\leqslant n$ and $\sum\limits_{k=1}^nx_{k,n}\leqslant b_n^2x_{n,n}$ where $b_n^2=\sum\limits_{k=1}^nm_k^2$. If $x_{n,n}\to0$, this yields $\sum\limits_{k=1}^nx_{k,n}\ll b_n^2$, which is Lindeberg condition.
But $\mathrm E((X-1)^2:X\geqslant t)=(t^2+1)\mathrm e^{-t}$ for every $t\geqslant0$, hence $x_{n,n}=O(b_n^2\mathrm e^{-\varepsilon b_n/m_n})$, which is enough to prove that $x_{n,n}\to0$.
