<s>[INST] Prove that $\left |\frac{1}{\sqrt{n}} \sum_{i=1}^{[tn]}X_i-\frac{\sqrt{t}}{\sqrt{[tn]}} \sum_{i=1}^{[tn]}X_i\right |\overset{n}{\to}0$ in probability 
As the title states, I would like to prove that
$$\left | \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor tn \rfloor }X_i  - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \sum_{i=1}^{\lfloor tn \rfloor}X_i \right |$$
converges to zero, with $n $, in probability. Here $X_i $ are independent random variables with zero mean and second moment equal to one. $\lfloor x \rfloor $ denotes the floor function and $t $ is a positive real number.


My first thought was that
$$\left | \frac{1}{\sqrt{n}} \sum_{i=1}^{\lfloor tn \rfloor}X_i  - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \sum_{i=1}^{\lfloor tn \rfloor}X_i \right |=\left |\sum_{i=1}^{\lfloor tn \rfloor}X_i \right |\left| \frac{1}{\sqrt{n}}   - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \right |$$
Where $\left| \frac{1}{\sqrt{n}}   - \frac{\sqrt{t}}{\sqrt{\lfloor tn \rfloor}} \right | \overset {n } {\to } 0 $. But does it converge sufficiently fast so that the product with $\left |\sum_{i=1}^{\lfloor tn \rfloor}X_i \right | $ converges to zero?
Most grateful for any help provided!
 [/INST] Here's an attempt to an answer using the points from stochasticboy321's comment above.
We will try to apply Chebyshev's inequality which in our case states, since the variance of $\sum_{i=1 } ^{\lfloor nt \rfloor } X_i$ is $\lfloor nt \rfloor$, that
$$P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i | \ge \epsilon \sqrt{\lfloor nt \rfloor}] \le \frac{1}{\epsilon^2}, \ \epsilon>0$$
for which is would be sufficient to show that, for some $n_0 $ onwards,
$$\left | \sqrt {\frac{t}{\lfloor nt \rfloor}}   - \frac{1}{\sqrt {n}}  \right |  \le \left | \frac{1}{tn} \right |$$
since then we would have [for $n \ge n_0 $]
$$P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i |\left | \frac{1}{\sqrt{n}} - \frac{\sqrt{t}}{\sqrt{\lfloor nt \rfloor}} \right | \ge \epsilon ] \le P[|\sum_{i=1 } ^{\lfloor nt \rfloor } X_i |\left | \frac{1}{\lfloor tn \rfloor} \right | \ge \epsilon ] \le \frac{1}{\epsilon^2 \lfloor tn \rfloor} \overset {n } {\to}  0$$

Attempt to show that $ \left | \frac{1}{\sqrt{n}} - \frac{\sqrt{t}}{\sqrt{\lfloor nt \rfloor}} \right | = \mathcal{O}(t^{-1 } n^{-3/2 })$:
From the inequality $ nt -1 < \lfloor nt \rfloor  \le nt $ we get 
$$\frac{1}{nt} \le \frac{1}{\lfloor nt \rfloor} < \frac{1}{nt-1} \implies \frac{1}{n} \le \frac{t}{\lfloor nt \rfloor} < \frac{t}{nt-1}=\frac{1}{n} + \frac{1}{n(nt-1)}$$
and since $0 \le a \le b < c \implies b-a \le c-a $ we get 
$$\frac{1}{n} - \frac{t}{\lfloor nt \rfloor} \le \left (\frac{1}{n} + \frac{1}{n(nt-1)}\right ) - \frac{1}{n} = \frac{1}{n(nt-1)}$$
and thus
$$\left |\frac{t}{\lfloor nt \rfloor}  - \frac{1}{n} \right | \le \left | \frac{1}{n(nt-1)} \right |$$
Since $|\sqrt{a} -  \sqrt{b}| \le \sqrt{|a-b|}$ holds for any two non negative numbers [see eg here] we get the inequality.
$$\left | \sqrt {\frac{t}{\lfloor nt \rfloor}}   - \frac{1}{\sqrt {n}}  \right | \le \left | \frac{1}{\sqrt {n(nt-1)}} \right |$$
I don't think it is true that
$$\frac{1}{\sqrt {n(nt-1)} } = \mathcal{O}(t^{-1 } n^{-3/2})$$
but I do believe we have that
$$\frac{1}{\sqrt {n(nt-1)} } = \mathcal{O}(t^{-1 } n^{-1})$$
which as noted above would be sufficient.

Any comment would be much appreciated!
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