Almost sure convergence of Bernoulli r.v. Assume we have an independent sequence of Bernoulli r.v. $(X_n)_{n=1}^\infty$
each $X_n$ gets 1 w.p. $\frac{1}{\sqrt{n}}$ and otherwise $0$.
How can we show that almost surely the following holds? 
$$\lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^N X_n  = 0  $$
Let's start with a sequence that is known to converge $a.s.$:

Let $B_n(p)$ be a sequence of iid Bernoulli r.v.s with $P(B_n(p)=1)=p$

In this case, it is known that 
$$\lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^N B_n(p) = p\; (a.s.)$$

For a given $p \in (0,1]$, we can let $M_p=\lceil \frac{1}{p^2}\rceil$ and we get:

$$\lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n \leq \lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^N B_n(p) = p \;(a.s.)$$

This is due to the fact that both $X_n$ and $B_n(p)$ are Bernoulli, but $X_n$ has a strictly decreasing probability of being $1$. This means that $\frac{1}{N}\sum _{n=M_p}^{N+M_p} X_n$ is bounded in probability by $\frac{1}{N} \sum _{n=1}^N B_n(p)$; i.e., 

$$P\left(\frac{1}{N}\sum_{n=M_p}^{N+M_p} X_n &gt; 0\right) \leq P\left(\frac{1}{N} \sum_{n=1}^N B_n(p) &gt; 0\right)$$

From the above two arguments, we have:

$$\forall p \in (0,1]\;\;\exists M_p\in \mathbb{N}: \lim_{N\rightarrow \infty} \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n \leq p\; (a.s.) $$

So, each value of $p \in (0,1]$ represents an almost-sure upper bound on the limit of an appropriately initialized series in $X_n$ (i.e., a suitably large $M_p$).

What about the $M_p-1$ terms  we omit from $\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n$. We know that these terms can add up to at most $M_p-1$, so our revised upper bound is:

$$\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^{N} X_n \leq \lim\limits_{N\rightarrow \infty} \left[ \frac{M_p-1}{N} +  \frac{1}{N} \sum _{n=M_p}^{N+M_p} X_n \right] \leq 0+p=p\; a.s.$$

We've shown that each $p \in (0,1]$ represents an almost-sure upper bound on the value of $\lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^{N} X_n $

Recasting in terms of the definition of limits, we can see that:

$$\forall p \in (0,1], \exists M: \left|\frac{1}{N} \sum _{n=1}^{N} X_n\right|&lt;p\;\forall N&gt;M\; (a.s.) \implies \lim\limits_{N\rightarrow \infty} \frac{1}{N} \sum _{n=1}^{N} X_n = 0 \; a.s.$$
$R_n=\frac{1}{N} \sum _{n=1}^N X_n=\frac{1}{N}(\sum _{n=1}^\sqrt{N}{X_n})+\frac{1}{N}(\sum _{n=\sqrt{N}+1}^N{X_n})$

the first summand is $ \leq \frac{\sqrt(N)}{N}$ (number of terms)

the second summand has less than N terms each of them having an expectation smaller than $\frac{1}{\sqrt{N}}$. As it is divided by N, its expectation is smaller than $\frac{1}{\sqrt{N}}$

So the expectation of $R_n$ is $\leq \frac{2}{\sqrt{N}}$