# Upper bound and monotony

I'm trying to figure out wether I understand the meaning of the following sequence correctly. Let $\displaystyle a_n = \sum_{k=1}^n\frac{1}{k+n}, n \in \mathbb{N}$.

Is this a correct upper bound?

$$\sum_{k=1}^n\frac{1}{k+n} < \sum_{k=1}^n\frac{1}{n} = n \cdot\frac{1}{n} = 1.$$

Is there a way to proof the monotony?

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A duplicate? $$a_{n+1}-a_n=\frac1{2n+2}+\frac1{2n+1}-\frac1{n+1}$$ – Jyrki Lahtonen Nov 20 '11 at 18:25

## 1 Answer

Since for all $n\in\mathbb N$ \begin{align*}a_{n+1}-a_n&=\frac 1{2n+2}+\sum_{k=1}^n\left(\frac 1{n+k+1}-\frac 1{n+k}\right)\\ &=\frac 1{2n+2}+\frac 1{2n+1}-\frac 1{n+1}\\ &=\frac 1{2n+1}-\frac 1{2n+2}\geq 0, \end{align*} the sequence $\{a_n\}$ is strictly increasing.

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