stochastic processes - Show that $X_ninmathcal{H}$, where $mathcal{H}:={h(t):h(t)text{ is an adapted process, }mathbb{E}[int_0^{infty}h^2(t)dt]

I am not sure if I got this exercise right... I have 2 questions:

1. Have I obtained the final result correctly?


2. If so, I used Wolfram Alpha to obtain the value of the series, but how else can I obtain $\sum_{k=0}^{n-1}\frac{k}{n^2}=\frac{n-1}{2n}$ using my own calculations?

Thanks a lot for your help!

QUESTION:

Let $W(t)$, $t\in\mathbb{R}_+$ be a Brownian motion with its natural filtration $\mathcal{F}_t, t\in\mathbb{R}_+$. Let
$$\mathcal{H}:=\{h(t):h(t)\text{ is an adapted process, }\mathbb{E}\left[\int_0^{\infty}h^2(t)dt\right]<\infty\}$$
denote the set of general integrands with respect to $W(t)$.

Consider the stochastic processes
$$X_n(t):=\sum_{k=0}^{n-1}W\left(\frac{k}{n}\right)\mathbb{1}_{\left(\frac{k}{n},\frac{k+1}{n}\right]}(t), t\ge 0,$$
for $n\ge 1$ and define $X(t):=W(t)\mathbb{1}_{[0,1]}(t),t\ge 0$.

Q) Verify that $X_n\in\mathcal{H}$ for all $n\ge 1$. (You may use Fubini's theorem without its proof)

ATTEMPT:

We have, using Fubini's theorem,
$$\mathbb{E}\int_0^{\infty}X_n^2(t)dt=\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\mathbb{E}\left(W\left(\frac{k}{n}\right)\right)^2dt=\sum_{k=0}^{n-1}\frac{k}{n}\int_{\frac{k}{n}}^{\frac{k+1}{n}}1 dt=\sum_{k=0}^{n-1}\frac{k}{n}\left[t\right]_{\frac{k}{n}}^{\frac{k+1}{n}}=\sum_{k=0}^{n-1}\frac{k}{n}\left(\frac{1}{n}\right)=\sum_{k=0}^{n-1}\frac{k}{n^2}=\frac{n-1}{2n}<\infty$$