I am working an exercise stated as follows:
Let $S_{n}=\sum_{k=1}^{n}Y_{k}$ for independent random variable $(Y_{k})$ such that $Var(Y_{k})<B<\infty$ and $\mathbb{E}Y_{k}=0$ for all $k$. Show that $$\dfrac{S_{n}}{\sqrt{n(\log n)^{1+\epsilon}}}\longrightarrow 0\ \text{a.s.}$$
My idea was to use Krnonecher's lemma that if $0<a_{n}\nearrow\infty$, and $\sum_{n=1}^{\infty}x_{n}/a_{n}<\infty$, then $a_{n}^{-1}\sum_{k=1}^{n}x_{k}\longrightarrow 0$ a.s.
So we replace $a_{n}=\sqrt{n(\log n)^{1+\epsilon}}$ which clearly $\nearrow \infty$, and replace $X_{k}=Y_{k}$, it then reminds to show that $$\sum_{k=1}^{\infty}\dfrac{Y_{k}}{a_{k}}<\infty.$$ To show this, my idea was to use Kolmogorov's 1 series.
Firstly, $\mathbb{E}Y_{k}/a_{k}=0$ and $Var(Y_{k}/a_{k})\leq B/a_{k}^{2}.$
However, the problem arose when I tried to compute the infinite sum of variance.
You could see the problem here is that $$\sum_{k=1}^{n}Var(Y_{k}/a_{k})\leq B\sum_{k=1}^{\infty}\dfrac{1}{k(\log k)^{1+\epsilon}},$$ but the summand has singularity at $k=1$, so we can only say this sum is finite if we start at $k=2$. But then we lost the power to use Kolmogorov.
On the other hand, we are also not satisfying the condition that $a_{n}>0$, since $a_{1}=0$, which is exactly what causes us the problem.
Is there anyway for me to modify a little bit so that my proof can go through?
It is perfectly fine if you point out that this exercise is not doable :)
Thank you!
Edit 1:
As Kabo Murphy pointed out, we can modify the sequence a little bit without affecting the limit. However, we do not need to modify $Y_{k}$ as it is not where our problem comes from.
We only need to set $a_{1}$ to be anything between $0$ and $a_{2}$, that is $0<a_{1}\leq a_{2}$, so that $a_{n}$ is a positive increasing sequence. And that's it, since $a_{1}$ then will not create $0$ term in the infinite sum of variance, so we just split the sum into $k=1$, and $k\geq 2$, which definitely converges.
Then the result follows from Kronecker's lemma, even though $a_{1}$ does not admit the general formula of $a_{n}$ for $n\geq 2$, since we take $n\longrightarrow\infty$ (so $a_{1}$ is not really important).
The term $k=1$ can easily be ignored (without changing the $a_n$'s). Let $Y_1'=0$ and $Y_i'=Y_i$ for $i >1$. If you have shown that the result holds for this new sequence you get it for the original sequence by the simple fact that $\frac 1 {\sqrt {n(\log n)^{1+\epsilon}}} \to 0$.