Sequence of consecutive differences $x_{k+1}-x_k$ square summable implies bounded sequence $\{x_k\}_{k\in\mathbb{N}}$.

Question

Consider a sequence of vectors $x_k\in\mathbb{R}^n$ and the sequence of squared consecutive differences $\{z_k\}_{k\in\mathbb{N}} = \{\|x_{k+1}-x_k\|^2\}_{k\in\mathbb{N}}$. What can we say about the boundedness of the sequence $\{x_k\}_{k\in\mathbb{N}}$? Of course, if it was summable outright then we would have convergence of $\{x_k\}_{k\in\mathbb{N}}$ by a triangle inequality argument. I tried to do something with

$$\|x-y\|^2\leq 2\|x\|^2 + 2\|y\|^2$$

but then you get a growing power of $2$ and you cannot use the square summability anymore.

Answer 1

If $x_n=1+\frac12+\ldots +\frac1n$, then $x_n\to\infty$, but $\sum |x_{n}-x_{n-1}|^2=\sum\frac 1{n^2}=\frac {\pi^2}6$.