Convergence in probability Let $(\xi_n)$ be a sequence of i.i.d. random variables with mean $0$ and variance $\sigma^2 < \infty$ and $s>0$. I need to prove that 
$$\frac{1}{\sqrt{\lfloor ns \rfloor}} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i -\frac{1}{\sqrt{ ns} } \sum_{i=1}^{\lfloor ns \rfloor} \xi_i  \stackrel{\mathbb{P}}{\longrightarrow} 0$$
Where the convergence above is in probability.
I tried to use the markov inequality to bound $$\mathbb{P} \left( \left|\frac{1}{\sqrt{\lfloor ns \rfloor}} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i -\frac{1}{\sqrt{ ns} } \sum_{i=1}^{\lfloor ns \rfloor} \xi_i  \right|>\delta \right)$$
but I didn't have success.
Any help will be appreciated.
 A: For your specific problem, we know that $\forall n\exists \varepsilon_n: ns=\lfloor ns \rfloor+\varepsilon_n,0\leq \varepsilon_n \leq1\; $ Which makes your problem equivalent to:
$$\frac{1}{\sqrt{\lfloor ns \rfloor}} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i -\frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \sum_{i=1}^{\lfloor ns \rfloor} \xi_i  \stackrel{\mathbb{P}}{\longrightarrow} 0$$
We can also pull out the sums:
$$\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)\stackrel{\mathbb{P}}{\longrightarrow} 0$$
Since we can bound $\varepsilon_n$, we can bound the absolute value of the LHS between 0 and another sequence:
$$0\leq \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)\right| \leq \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \right|=\left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right| \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right)$$
We can pull out $$\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }}$$ since it is non-negative.
We can also see that:
$$\left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \to 0^+ \;\;(\text{surely})$$
Now, lets bound the rate of convergence:
$$\left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \leq \frac{1}{\lfloor ns \rfloor}$$

Proof
Let $k_n>0$ be a monotonic increasing, unbounded sequence.
$$ k_n\left(\frac{1}{\sqrt{k_n}} - \frac{1}{\sqrt{k_n+1 }} \right) = \sqrt{k_n}-\frac{k_n}{\sqrt{k_n+1}}$$
$$\lim_{k_n\to \infty} \sqrt{k_n}-\frac{k_n}{\sqrt{k_n+1}} = 0 \implies \frac{1}{\sqrt{k_n}} - \frac{1}{\sqrt{k_n+1 }} = o\left(\frac{1}{k_n}\right)$$

Using this convergence, we can see that:
$$ \forall s>0,\exists c \in  \mathbb{N}: \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right| \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+1 }} \right) \leq \left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right|\frac{1}{ns}\;\forall n>c$$
Also,
$$\left|\sum_{i=1}^{\lfloor ns \rfloor} \xi_i\right|\frac{1}{ns} \leq \sum_{i=1}^{\lfloor ns \rfloor} \left|\xi_i\right|\frac{1}{ns}\stackrel{\mathbb{P}}{\longrightarrow}  E[\left|\xi_i\right|] <\infty$$
Since $\xi_i$ have finite mean and variance.
So, we know that the sequence$\left\{\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)\right\}_n$ converges to a some finite value $T:|T|\leq\sigma^2<\infty$ in probability.

We will need to be more specific, since our first bounding was too coarse and only established that the series does not diverge in probability. Lets look at the variance of:
$$G_n:=\sum_{i=1}^{\lfloor ns \rfloor} \xi_i \left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)$$
$$Var(G_n)=\sigma^2_n=\lfloor ns \rfloor\sigma^2\left(\frac{1}{\sqrt{\lfloor ns \rfloor}} - \frac{1}{\sqrt{\lfloor ns \rfloor+\varepsilon_n }} \right)^2$$
From our earlier arguments, we can see that $\sigma^2_n\to 0$.
Also, we can see that $\forall n, E[G_n]=0$. Therefore, by Chebyshev's theorem:
$$P(|G_n-0|\geq k\sigma^2_n)\leq \frac{\sigma^2_n}{k^2} \to 0 \;\;\forall k>0 \implies G_n \stackrel{\mathbb{P}}{\longrightarrow} 0$$
