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!

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!
