Let $(E,d)$ be a metric space and $(x_n)_{n\in\mathbb N}\subseteq E$. We can easily show that if $(\varepsilon_n)_{n\in\mathbb N}\subseteq[0,\infty)$ is summable and $$d(x_n,x_{n+1})\le\varepsilon_n\;\;\;\text{for all }n\in\mathbb N,\tag1$$ then $(x_n)_{n\in\mathbb N}$ is Cauchy.
Now let $H$ be a Hilbert space, $(x_n)_{n\in\mathbb N_0}\subseteq H$ and $h_n:=x_n-x_{n-1}$. Assume $(h_n)_{n\in\mathbb N}$ is orthogonal and $\sum_{n\in\mathbb N}\left\|h_n\right\|_H^2<\infty$.
Are we able to conclude that $(x_n)_{n\in\mathbb N}$ is Cauchy?
We cannot apply the former result for general metric spaces, since we've only got $\sum_{n\in\mathbb N}\left\|h_n\right\|_H^{\color{red}2}<\infty$. On the other hand, this implies that $(h_n)_{n\in\mathbb N}$ is summable. Is this enough to obtain that $(x_n)_{n\in\mathbb N}$ is Cauchy?
Yes. Let $\varepsilon>0$ and find $n_0\in\mathbb{N}$ such that $\sum_{j=n_0+1}^\infty\|h_j\|^2<\varepsilon^2$. Now let $m>n\ge n_0$. We have $x_m-x_n=x_m-x_{m-1}+x_{m-1}-x_{m-2}+\dots+x_{n+1}-x_n=\sum_{j={n+1}}^{m}h_j$. Therefore $$\|x_n-x_m\|^2=\bigg\|\sum_{j=n+1}^mh_j\bigg{\|}^2=\sum_{j=n+1}^m\|h_j\|^2\le\sum_{j=n_0+1}^\infty\|h_j\|^2<\varepsilon^2$$ so $\|x_n-x_m\|<\varepsilon$.