Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence. 
Let $n\in \mathbb N$ and:
  $$
x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n
$$
  Show that $\{x_n\}$ is a bounded sequence.

This sequence appears a bit tricky because it involves harmonic series. Below are steps I take.
Lower bound:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \ge \sqrt{\sum_{k=1}^n\left(\frac{k}{k}\right)^2} = \sqrt{\sum_{k=1}^n1}=\sqrt{n}\implies\\
\implies x_n \ge \sqrt n - \sqrt n \ge 0
$$
Lower bound is simple. 
Upper bound:
To get rid of radical lets use Cauchy-Schwarz (note the below is incorrect as shown in Zvi's answer):
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \le \sqrt{\left(\sum_{k=1}^n\left(\frac{k+1}{k}\right)\right)^2} =\sum_{k=1}^n\left(\frac{k+1}{k}\right) = n+\sum_{k=1}^n{1\over k} = n + H_n
$$
So this doesn't show $x_n$ is bounded above. I've tried another approach:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}
$$
Consider nominator:
$$
\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=n+\sum_{k=1}^n{2\over k}+\sum_{k=1}^n{1\over k^2}-n = \sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}
$$
For denominator:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n = \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n
$$
So $x_n$ is:
$$
x_n = \frac{\sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}}{ \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n }
$$
But i don't see how to proceed from this point. What else could i try? How to show $x_n$ is bounded above?
Please note the precalculus tag.

Even if $\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n + 2H_n + H_n^{(2)}$ involves (generalized) harmonic numbers, it is pretty straightforward to notice$^{(*)}$ that $H_n=O(\log n)$ and $H_{n}^{(2)}=O(1)$ as $n\to +\infty$, hence
$$ \sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2 = n\left(1+O\left(\frac{\log n}{n}\right)\right),\tag{1}$$
$$ \sqrt{\sum_{k=1}^{n}\left(1+\frac{1}{k}\right)^2} = \sqrt{n}\left(1+O\left(\frac{\log n}{n}\right)\right)=\sqrt{n}+O\left(\frac{\log n}{\sqrt{n}}\right)\tag{2}$$
and the wanted limit is zero.
$(*)$ $H_n \leq 1+\int_{1}^{n}\frac{dx}{x}=1+\log(n)$ and $H_{n}^{(2)}\leq 1+\sum_{k=1}^{n}\frac{1}{k(k+1)}\leq 2$.
