# 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?

• Side-note: The sequence converges to $\ln 2$ – Jakobian Oct 15 '18 at 15:45

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

$$x_n \leq \sum_{k=1}^{n}\frac{1}{\sqrt{k+n}\sqrt{k+n-1}}\stackrel{\text{Cauchy-Schwarz}}{\leq}\sqrt{n\sum_{k=1}^{n}\left(\frac{1}{k+n-1}-\frac{1}{k+n}\right)}=\frac{1}{\sqrt{2}}.$$

$$x_n = \sum\limits_{k=1}^{n} {1\over n+k}=\sum\limits_{k=1}^{n} \frac{1}{n}{1\over 1+\frac{k}{n}}=\frac{1}{n} \sum\limits_{k=1}^{n}{1\over 1+\frac{k}{n}}$$ Riemann sum $$=\int\limits_0^1 \frac{1}{1+x}dx=\ln2$$