Strong law of large numbers for square-integrable and uncorrelated random variables with bounded variance Let $(\Omega,\mathcal{A},P)$ be a probability space and  $(X_n)_{n\in\mathbb{N}}$ be a sequence of square-integrable and uncorrelated (maybe we actually need independence) random variables $\Omega\to [0,\infty]$ with $V:=\sup_{n\in\mathbb{N}}\operatorname{Var}X_n<\infty$.
Moreover, let $$X^{(n)}:=\sum_{i=1}^nX_i\;\;\;\text{and}\;\;\;\overline{X}^{(n)}:=\frac{1}{n}X^{(n)}$$
I want to show, that it holds: $$\limsup_{i\to\infty}\left|\overline{X}^{(i)}-E\left[\overline{X}^{(i)}\right]\right|=0\;\;\;\text{almost surely}\tag{1}$$
Proof: Let $\varepsilon>0$ and $$k_n:=\lfloor(1+\varepsilon)^n\rfloor\ge\frac{1}{2}(1+\varepsilon)^n\tag{2}$$
Chebyshev's inequality yields
\begin{equation}
\begin{split}
\sum_{n\in\mathbb{N}}\Pr\left[\left|\overline{X}^{(k_n)}-E\left[\overline{X}^{(k_n)}\right]\right|\ge\frac{1}{(1+\varepsilon)^{n/4}}\right]&\le&\sum_{n\in\mathbb{N}}(1+\varepsilon)^{n/2}\operatorname{Var}\overline{X}^{(k_n)}\\
&\stackrel{(*)}{\le}&\sum_{n\in\mathbb{N}}(1+\varepsilon)^{n/2}\frac{V}{k_n}\\&\stackrel{\text{(2)}}{\le}&2V\sum_{n\in\mathbb{N}}\frac{1}{(1+\varepsilon)^{n/2}}&<\infty
\end{split}
\end{equation}
where $(*)$ holds by uncorrelatedness and the formula of Bienaymé.
Thus, the lemma of Borel-Cantelli yields $$\Pr\left[\limsup_{n\to\infty}\left\{\left|\overline{X}^{(k_n)}-E\left[\overline{X}^{(k_n)}\right]\right|\ge\frac{1}{(1+\varepsilon)^{n/4}}\right\}\right]=0,$$ i.e. there is a $n\in\mathbb{N}$, such that for all $m\ge n$ $$\left|\overline{X}^{(k_n)}-E\left[\overline{X}^{(k_n)}\right]\right|<\frac{1}{(1+\varepsilon)^{n/4}}\;\;\;\text{almost surely}$$ So, we've got $$\limsup_{n\to\infty}\left|\overline{X}^{(k_n)}-E\left[\overline{X}^{(k_n)}\right]\right|=0\;\;\;\text{alsmost surely}\tag{3}$$ Now, observe that $$k_{n+1}\le(1+2\varepsilon)k_n$$ which implies that 
\begin{equation}
\begin{split}
\frac{1}{1+2\varepsilon}\overline{X}^{(k_n)}&\le& \frac{k_n}{k_{n+1}}\overline{X}^{(k_n)}\\
&=&\frac{1}{k_{n+1}}X^{(k_n)}\\&\le& \overline{X}^{(i)}\\&\le&\frac{1}{k_n}X^{(k_{n+1})}\\&\le &(1+2\varepsilon)\overline{X}^{(k_{n+1})}
\end{split}\tag{4}
\end{equation}
for all $i\in [k_n,k_{n+1}]\cap\mathbb{N}$.
How can we conclude, that $$L:=\limsup_{i\to\infty}\left|\overline{X}^{(i)}-E\left[\overline{X}^{(i)}\right]\right|=0\tag{5}$$ almost surely?
My first idea was to use $(4)$ to see that $$L\le\limsup_{n\to\infty}\left|(1+2\varepsilon)\overline{X}^{(k_{n+1})}-E\left[\frac{1}{1+2\varepsilon}\overline{X}^{(k_n)}\right]\right|\tag{6}$$However, I am not able to take advantage of $(3)$ in $(6)$.
 A: If you also include "uniformly bounded means," so that $E[X_n] \leq C$ for all $n$ (for some constant $C$) then you are okay. Take your last equation (4) (which is really equation (6)) and add and subtract the same thing to get terms $E[\overline{X}^{(k_n)}](1+2\epsilon) - E[\overline{X}^{(k_n)}](1+2\epsilon)$ to finish (via triangle inequality). 

I would also phrase your conclusion before (3) differently: "Almost surely, there exists an $n$ such that $|\overline{X}^{k_m}-E[\overline{X}^{k_m}]| < 1/(1+\epsilon)^{m/4}$ for all $m \geq n$. 

It seemed to me that taking away the "uniformly bounded means" assumption was very similar  to taking away the "non-negative" assumption, and so one should imply the other. Thus, at first I thought non-negativity (and hence "uniformly bounded mean") was crucial.  However, after playing around, I believe the following is a proof for the general "non-negative" case, which also does not require uniformly bounded means.  The proof uses existence of a sequence of non-decreasing positive integers $k_n$ that satisfy: 
1) $k_1 = 1$. 
2) $\lim_{n\rightarrow\infty} k_n = \infty$
3) $\sum_{n=1}^{\infty} 1/k_n < \infty$
4) $\sum_{n=1}^{\infty} \left(\frac{k_{n+1}}{k_n}-1\right)^2 < \infty$. 
Using $k_n = \left\lfloor \left(1 + \frac{1}{2n^{0.6}}\right)^n\right\rfloor$ will do, since $k_n \approx e^{(1/2)n^{0.4}}$ and so increases super-linearly, but $(k_{n+1}/k_n-1)^2 \approx \frac{1}{4n^{1.2}}$. 
Claim:  If $\{X_n\}_{n=1}^{\infty}$ are uncorrelated and $Var(X_n) \leq V$ for all $n$ (for some finite constant $V$), then (with prob 1): 
$$ \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n (X_n-E[X_n]) = 0 $$
Proof: Let $k_n$ be a non-decreasing sequence of positive integers with the above four properties.  Fix $\epsilon>0$.  By Chebyshev: 
$$ Pr\left[\left|\frac{1}{k_n}\sum_{i=1}^{k_n} (X_i-E[X_i])\right| \geq \epsilon  \right] 
\leq \frac{k_n V}{\epsilon^2 k_n^2}
= \frac{V}{\epsilon^2 k_n} $$
Thus, since $\sum_{n=1}^{\infty} 1/k_n < \infty$, we have: 
$$ \sum_{n=1}^{\infty} Pr\left[\left|\frac{1}{k_n}\sum_{i=1}^{k_n} (X_i-E[X_i])\right| \geq \epsilon  \right] < \infty $$
and so (with prob 1): 
$$\limsup_{n\rightarrow\infty} \left|\frac{1}{k_n}\sum_{i=1}^{k_n} (X_i-E[X_i])\right| \leq \epsilon $$ 
This holds for all $\epsilon>0$, and so (with prob 1): 
$$ \lim_{n\rightarrow\infty} \frac{1}{k_n}\sum_{i=1}^{k_n} (X_i-E[X_i]) = 0 \: \: \: (*) $$ 
Now fix $m \in \{1, 2, 3, \ldots\}$ and define $n_m$ as the smallest index such that $k_{n_m} \leq m \leq k_{n_m+1}$.  Then: 
\begin{align} 
&\frac{1}{m}\sum_{i=1}^m (X_i-E[X_i])\\
&=\frac{k_{n_m+1}}{m}\frac{1}{k_{n_m+1}}\sum_{i=1}^{k_{n_m+1}} (X_i-E[X_i])  - \frac{1}{m}\sum_{i=m+1}^{k_{n_m+1}} (X_i-E[X_i])
\end{align} 
If we can show both terms on the right-hand-side go to 0 as $m\rightarrow \infty$, we are done.  Taking a limit as $m\rightarrow \infty$ of the first term on the right-hand-side gives 0 (with prob 1) by equation (*). It suffices to prove that (with prob 1): 
$$ \lim_{m\rightarrow\infty}  \frac{1}{m} \sum_{i=m+1}^{k_{n_m+1}}(X_i-E[X_i])=0 $$
To this end, fix $\delta>0$. We have by Chebyshev: 
\begin{align} 
&Pr\left[\left|\frac{1}{m} \sum_{i=m+1}^{k_{n_m+1}}(X_i-E[X_i])\right| \geq \delta\right] \\
& \leq \frac{V(k_{n_m+1}-m)}{m^2\delta^2} \\
& \leq \frac{V(k_{n_m+1}-k_{n_m})}{k_{n_m}^2 \delta^2} 
\end{align} 
Summing over $m$ gives: 
\begin{align} 
&\sum_{m=1}^{\infty} Pr\left[\left|\frac{1}{m}\sum_{i=m+1}^{k_{n_m+1}}(X_i-E[X_i])\right| \geq \delta\right] \\
&\leq \frac{V}{\delta^2}\sum_{m=1}^{\infty} \frac{(k_{n_m+1}-k_{n_m})}{k_{n_m}^2}\\
&\leq \frac{V}{\delta^2}\sum_{n=1}^{\infty}\frac{(k_{n+1}-k_n)^2}{k_{n}^2} \\
&= \frac{V}{\delta^2}\sum_{n=1}^{\infty} \left(\frac{k_{n+1}}{k_n}-1\right)^2 \\
&< \infty
\end{align} 
Thus, with prob 1: 
$$ \limsup_{m\rightarrow\infty} \left|\frac{1}{m} \sum_{i=m+1}^{k_{n_m+1}}(X_i-E[X_i])\right| \leq \delta $$
This holds for all $\delta$, and so the $\limsup$ must be 0 (with prob 1). 
