$\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $ for equally spaced interval?

Question

{Q: [0, T] is spaced equally, as usual notation:$\Delta B_i=B_{t_i}-B_{t_{i-1}}$, $t_i-t_{i-1}=T/n$. Show $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $}

$Proof$: for all $\varepsilon$,

$\mathbb{P}\left(\{\omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-t|>\varepsilon\}\right)$

$\leq {\frac{E(\sum_{i=1}^{n}\Delta B_i^2-t)^4}{\varepsilon^4}}$(by Markov inequality)

$=12T^4(1/n^2+4/n^4)=O(1/n^2)$ .

then $\sum \mathbb{P}\left(\{ \omega\in \Omega:|\sum_{i=1}^{n} \Delta B_i^2-T|>\varepsilon\}\right)$ is a convergent series, satisfying the definition of c.c. (complete convergence),

So $\sum_{i=1}^{n} \Delta B_i^2\stackrel{ c.c. } \longrightarrow T $, that also implies $a.s.$ (or by Borel-Cantelli lemma).

Is there any error I cannot find? so it is appreciated if someone can point it out. wellcome for any comments, thanks so much.