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.

You have, using that $a^2-b^2=(a+b)(a-b)$, 
\begin{align}
\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=\sum_{k=1}^n \left[ \left(1+\frac1k\right)^2-1\right]
=\sum_{k=1}^n \frac1k\left(2+\frac1k\right)\leq3\sum_{k=1}^n\frac1k.
\end{align}
Using the comparison with the integral we get 
$$
\sum_{k=1}^n\frac1k\leq1+\int_1^n\frac1t\,dt=1+\log n. 
$$
Then
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}\leq\frac{3(1+\log n)}{\sqrt n}.
$$
Since $\lim_{x\to\infty}\frac{3(1+\log x)}{\sqrt x}=0$, the continous function $f(x)=\frac{3(1+\log x)}{\sqrt x}$ is bounded on $[1,\infty)$. Thus there exists $C>0$ with 
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n \leq\frac{3(1+\log n)}{\sqrt n}\leq C. 
$$
