Validate the proof that the sequence $x_n = \sum_{k=1}^{n} {1\over n+k}$ is bounded. 
Let $n \in \mathbb N$ and:
  $$
x_n = \sum_{k=1}^{n} {1\over n+k}
$$
  Prove that $x_n$ is a bounded sequence.

I'm wondering whether the proof below is valid. 
Since $n \in \mathbb N$ we have that $x_n$ is strictly greater than $0$. For the upper bound lets consider the following sequence $y_n$:
$$
\begin{align}
y_n &= {1 \over n + 1} + {1 \over n + 1} + \dots + {1 \over n + 1} = \\
&= \sum_{k = 1}^n {1 \over n+1} = {n \over n + 1}
\end{align}
$$
Since $x_n$ has an increasing denominator in each consecutive term of the sum we may conclude that $x_n < y_n$. So summarizing the above:
$$
0 < x_n < y_n
$$
Which means that the sequence is bounded. Have I missed something?

Yes, you have $(\forall n\in\mathbb{N}):0<x_n<y_n$, but asserting that the sequence $(x_n)_{n\in\mathbb N}$ is bounded means that there are constants $a$ and $b$ such that$$(\forall n\in\mathbb{N}):a<x_n<b.$$That's easy, though, after what you did. Just take $a=0$ (of course) and $b=1$.
