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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?

  • 1
    [A duplicate?](http://math.stackexchange.com/q/73550/11619) $a_{n+1}-a_n=\frac1{2n+2}+\frac1{2n+1}-\frac1{n+1}$2011-11-20

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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.