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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?

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    $\begingroup$ Side-note: The sequence converges to $\ln 2$ $\endgroup$ – Jakobian Oct 15 '18 at 15:45
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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 $b=1$.

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

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

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