Does strong law of large numbers hold for an average of this triangular array containing "almost i.i.d." sequences? If a sequence of $n$ i.i.d. positive random variables $X_1,\ldots,X_n$ has the expectation $\mathbb{E}[X_1]=\mu<\infty$, then it is known that the strong law of large numbers (SLLN) holds: $\frac{1}{n}\sum_{i=1}^n X_i\xrightarrow{a.s.}\mu$, even when the second moment is not finite. 
I am wondering if it holds when the sequence contains one element with mean that increasing in $n$ (but not too fast).  In fact, I am asking a question about a triangular array.
In my particular case, I have a triangular array $\{\{X_{1n},\ldots,X_{(n-1)n}, X_{nn}$ where $X_{in}>0$ and independent across all rows and columns; $X_{1n},\ldots,X_{(n-1)n}$ are identical with $\mathbb{E}[X_{1n}]=1$ and $\mathbb{E}[X_{1n}^2]=e^{2(\log n)^{a}}$ where $0<a<1$; and $\mathbb{E}[X_{nn}]=e^{(\log n)^{a}}$ and $\mathbb{E}[X_{nn}^2]=e^{4(\log n)^{a}}$.  Since $\lim_{n\rightarrow\infty}\frac{e^{(\log n)^{a}}}{n}=0$, the contribution of $X_{nn}$ to the average $\frac{1}{n}\sum_{i=1}^n X_{in}$ is small, and, thus, maybe the SLLN holds "row-wise".  I can use the Chebyshev's inequality to prove the weak law of large numbers for this case.  However, my tool for proving almost sure convergence, the Borel-Cantelli lemma, seems to not work here at all (though I might not be using correctly).  Can someone help?
Minor edits to above: 
Per responses and comments, I corrected some typos as well as notation for the random variables, and clarified that what I am dealing is, in fact, a triangular array.
Main edit
I was asked in the comments/answers about where this question came from and more information about $X_i$'s.  Here I describe the setting from which this problem arose.
I am given a machine which takes $n$ as input.  As an output it gives me a sequence of $n$ random variables $X_{1n},\ldots,X_{nn}$, where:
$$X_{in}=\exp\left[\frac{f(n)\Gamma_{in}}{1+f(n)}-k\left(\frac{f(n)}{1+f(n)}+\left[\frac{f(n)}{1+f(n)}\right]^2\right)\right], \text{  when }i\neq S\\
X_{in}=\exp\left[f(n)\Gamma_i-k\left(\frac{f(n)}{1+f(n)}+\left[\frac{f(n)}{1+f(n)}\right]^2\right)\right], \text{  when }i=S\\$$
where $k=n^c$, $0<c<1$, the sequence $\Gamma_{1n},\ldots,\Gamma_{nn}$ contains $n$ i.i.d. chi-squared random variables, each with $k$ degrees of freedom (note, the number of degrees of freedom for each $\Gamma_{in}$ depends on $n$, however, the independence holds across rows), $f(n)=\sqrt{\frac{(\log n)^{a}}{k}}$, and $1\leq S \leq n$ is unknown.  I am interested in the average $\frac{1}{n}\sum_{i=1}^n X_{in}$ and, specifically would like to show that, even though I don't know what $S$ is, $X_{Sn}$ does not impact the average for large enough input $n$.  The statement I want is this: for any $\epsilon>0$, with probability one there exists $n_0$ such that for all $n\geq n_0$, $\left|\frac{1}{n}\sum_{i=1}^n X_{in}-1\right|<\epsilon$.
Using the formula for the moment generating function of chi-squared random variable, and the Taylor series expansion of $\log(1-x)$ around $x=0$, I arrive at the means:
$$\begin{array}{rcl}E[X_{in}]&=&\exp\left[\mathcal{O}\left(\sqrt{\frac{(\log n)^{3a}}{k}}\right)\right]\approx1, \text{ when }i\neq S\\
E[X_{Sn}]&=&\exp\left[-\frac{k}{2}\log\left(1-2f(n)\right)-k\left(\frac{f(n)}{1+f(n)}+\left[\frac{f(n)}{1+f(n)}\right]^2\right)\right]\\
&=&\exp\left[kf^2(n)+\mathcal{O}\left(\sqrt{\frac{(\log n)^{3a}}{k}}\right)\right]\\
&\approx&e^{(\log n)^a}
\end{array}$$
The second moments are computed similarly:
$$\begin{array}{rcl}E[X_{in}^2]&=&\exp\left[-\frac{k}{2}\log\left(1-\frac{4f(n)}{1+f(n)}\right)-2k\left(\frac{f(n)}{1+f(n)}+\left[\frac{f(n)}{1+f(n)}\right]^2\right)\right], \text{ when }i\neq S\\
&=&\exp\left[2kf^2(n)+\mathcal{O}\left(\sqrt{\frac{(\log n)^{3a}}{k}}\right)\right]\\
&\approx&e^{2(\log n)^a}\\
E[X_{Sn}^2]&=&\exp\left[-\frac{k}{2}\log\left(1-4f(n)\right)-2k\left(\frac{f(n)}{1+f(n)}+\left[\frac{f(n)}{1+f(n)}\right]^2\right)\right]\\
&=&\exp\left[4kf^2(n)+\mathcal{O}\left(\sqrt{\frac{(\log n)^{3a}}{k}}\right)\right]\\
&\approx&e^{4(\log n)^a}
\end{array}$$
Using Chebyshev's inequality, it's straightforward to show that for any $\epsilon>0$, $\delta>0$, there exists $n_0$ such that for all $n\geq n_0$,
$P\left(\left|\frac{1}{n}\sum_{i=1}^nX_i-1\right|>\epsilon\right)<\delta$, but I was hoping for a stronger result.  The discussion in answers and comments is discouraging, however.
 A: Edit: As Ahriman points out, the $X_i$ are not identical for $i<n$ since $E[X_i^2]$ depends on $n$. The answer I give only applies for $X_i$ identical for $i<n$.
Yes. By the usual SLLN, you have
$$\frac{1}{n-1}\sum_{i=1}^{n-1} X_i\overset{a.s.}{\longrightarrow}\mu$$
and since $\frac{n-1}{n}\to 1$ and $\frac{1}{n} X_n\to 0$ a.s., it follows that
$$\frac{1}{n}\sum_{i=1}^{n} X_i= \frac{1}{n}X_n + \frac{n-1}{n}\frac{1}{n-1}\sum_{i=1}^{n-1} X_i\overset{a.s.}{\longrightarrow} 0+1\cdot\mu=\mu.$$
A: (If you'd like a counterexample, I'll expand this to a full answer. Otherwise, please give more information about your particular sequence $X$.)
This doesn't quite hold without extra assumptions on the sequence $X$. Fix b=1, a=1/2. Let 
$$p_n = e^{-\sqrt{\log n}}$$ 
and 
$$ x_n = e^{2\sqrt{\log n}}.$$
Let $X_n$ be $x_n$ with probability $p_n$ and $0$ otherwise. $X_n$ has the appropriate mean and second moment. Now we'd like to show that the SLLN doesn't hold for this sequence. 
Fix $n\in\mathbb{N}$, sufficiently large. A little calculation shows that
$$ \mathbb{P}\left[\sum_{m=n(1-e^{-\sqrt{\log n}})}^nX_m \ge \frac n 2\right ] \ge \frac 1 2,$$
from which it follows that 
$$\limsup_n \frac {X_n} n \ge \frac 1 2,$$
a.s., by the second Borel-Cantelli Lemma. 
